Electronic Circuit Analysis
Second Edition
"This page is Intentionally Left Blank"
Electronic Circuit Analysis
Second Edition
Dr. K. Lal Kishore, Ph.D
Registrar,
Jawaharlal Nehru Technological University,
Kukatpally, Hyderabad - 500 072.
BSP BS Publications
4-4-309, Giriraj Lane, Sultan Bazar,
Hyderabad - 500 095 - A. P.
Phone: 040-23445688
Copyright © 2008, by Publisher
All rights reserved
I
No part of this book or parts thereof may be reproduced, stored in a
retrieval system or transmitted in any language or by any means,
electronic, mechanical, photocopying, recording or otherwise without
the prior written permission of the publishers.
Published by
BSpBS Publications
;;;;;;;;;= 4-4-309, Giriraj Lane, Sultan Bazar,
Hyderabad - 500 095 - A. P.
Phone: 040-23445688
e-mail: contactus@bspublications.net
www.bspublications.net
Printed at
Adithya Art Printers
Hyderabad.
ISBN:
81-7800-166-7
DEDICATED TO
Sri
Saraswati
the goddess of learning
"This page is Intentionally Left Blank"
PREFACE TO SECOND EDITION
Since publishing first edition of this book three years back, there are few additions in the
subject and also as a result of receiving some feedback, it has become imperative to bring
another edition to cover the lapses and bring the text mor:e useful to students.
In the second edition, I have reorganised the chapters and also added few subchapters
like High Frequency Amplifiers, Stability Considerations, UPS and SMPS in the respective
chapters.
The author is indebted to Sri. M.V. Ramanaiah, Associate Professor in the Department
of ECE, Gokaraju Rangaraju Institute of Engineering and Technology, Hyderabad for his
efforts in going through the book and making the symbols etc. more perfect which were
cropped up at the time of typing the text.
I am also thankful to Mr. Nikhil Shah and Mr. Manoj Jha of as Publications for their
persuasion and bringing the second edition ofthis book in record time.
-Author
"This page is Intentionally Left Blank"
PREFACE TO FIRST EDITION
Foundations for Electronics Engineering were laid as far back as 18 th Century when
H.A. Lorentz postulated the existence of negatively charged particles called as Electrons.
Since then the field of electronics engineering has developed rapidly. Advancement in this
area was more rapid since 1970s, with Digital Electronics dominating over Analog Electronics,
as was done by Solid State Devices in 1960s over Vacuum Tubes. After the Industrial
Revolution, it is Computer Revolution which is the astonishing phenomenon, at the fag end
of the 21 sl Century. The next striking development could be computer communications.
The research and development work done in the field of Semiconductor Devices and
Technology contributed significantly for the miniaturisation taking place in electronic systems
and computers. Thus Electronics Engineering is a fascinnating subject.
Electronic Circuit Analysis is an important component of the broad area of Electronics
and Communications Engineering. Electronic Circuit Design and Analysis aspects are dealt
with in this book. Learning these topics is very essential for any electronics engineer. A
student must study the subject, not just for the sake of passing the examination, but to learn
the concepts. In this competitive world, to secure a job or to learn the concepts, proper
effort must be made. This book is written WIth that motive. Any book written just for the
sake of enabling the student to pass the examination will not fullfil its complete objective.
Electronic Circuit Analysis is one of the fundamental subjects, which helps in I.C design,
VLSI design etc.
This textbook can also be used for M.Sc (Electronics), AMIETE, AMIE (Electronics)
B.Sc (Electronics), Diploma courses in Electronics, Instrumentation Engineering and other
courses where Electronics is one subject. So students from Universities, Engineering Colleges
and Polytechnics can use this book.
Though efforts are made to milllmize typing errors, printing mistakes and other
topographical errors, still, there could be some omissions. The author and publisher will be
thankful if such errors brought to notice for necessary correction.
x
Many Textbooks are referred while writing this book. The author is thankful for
them and their publishers.
The author is thankful to Mr. Nikhil Shah for the encouragement given to write this
book. The author is also thankful to Mr. Naresh, Mr. Prashanth, Mr. J. Das, Shri Raju and
other staff of MIS. B.S. Publications. The author is highly grateful to Prof. D. S. Murthy
Head, ECE Dept., Gayatri Vidya Parishad College of Engineering, Vizag, for his valuable
suggestions. The author is also thankful to Mrs. Mangala Gowari Assoc. Prof. Dept. of
ECE, JNTU, Hyderabad Mr. P. Penchalaiah Assoc. Prof. Dept of ECE, Vignan Inst. of
Science and Technology, Hyderabad and Mr. P. Ramana Reddy, APECE, JNTU CE,
Hyderabad. Author is particularly thankful to Mr. P. Penchalaiah for his effort in minimizing
printing mistakes. The author is also thankful to Ms. U.N.S. Sravanthi and Ms. Srujana for
the proof reading work.
For a task like writing textbook, there is always scope for improvement and corrections.
Suggestions are welcome.
~ffi ~~cp cft~ur:r:
~~Tfcf~ 3l~(fJ:f:
Looking in right perspective is wisdom.
Education gives prosperity.
Knowledge shows the path -
July 2003
Wisdom lights it up.
-Author
CONTENTS
Preface to Second Edition .................................................................................................. vii
Preface to First Edition ...........................................................: ........................................... ix
Symbols ................................................................................................................................ xv
Brief History of Electronics .............................................................................................. xix
Unit - 1
Single Stage Amplifiers ..................................................... 1
1.1
12
1.3
1.4
1.5
1.6
1.7
1.8
1.9
Introduction ........................................................................................................................... 2
Small Signal Analysis of Junction Transistor ......................................................................... 3
Common Emitter Amplifier ...................................................................................................... 4
Common Base Amplifier ......................................................................................................... 9
Common Collector Amplifier ................................................................................................ 12
JFET Amplifiers .................................................................................................................... 21
Common Drain (CD) Amplifier .............................................................................................. 27
Common Gate Amplifier (CG) ................................................................................................ 27
Gain - Bandwidth Product .................................................................................................... 29
Objective Type Questions .......................................................................................................... 30
Essay Type Questions .............................. ,.................................................................................. 31
Answers to Objective Type Questions ........................................................................................ 32
Unit - 2
Multistage Amplifiers ...................................................... 33
2.1
Multistage Amplifiers Methods of Inter Stage Coupling ...................................................... 34
22 n - Stage Cascaded Amplifier ................................................................................................ 43
2.3 Equivalent Circuits ................................................................. '" ........................................... 54
2.4 Miller's Theorem .................................................................................................................. 55
2.5 Frequency Effects ................................................................................................................ '5l
xii
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
Frequency Effects ................................................................................................................ 57
Amplifier Analysis ................................................................ ,............................................... 59
High Input Resistance Transistor Circuits ........................................................................... 65
The Cascode Transistor Configuration ................................................................................ 72
CE-CCAmplifiers ............................................................................................................... 77
Two Stage RC Coupled JFET amplifier (in Common Source (CS) configuration) .................. 78
Difference Amplifier .............................................................................................................. 78
Circuit for Differential Amplifier ............................................................................................ ll)
Objective Type Questions .............................................................................................................. 84
Essay Type Questions ................................................................................................................... 86
Answers to Objective Type Questions .......................................................................................... K7
Unit - 3
High Frequency Transistor Circuits ............................... 89
3.1 Transistors at High Frequencies .......................................................................................... 90
32 Hybrid - 1t Common Emitter Transconductance Model ....................................................... 90
3.3 Determination of Hybrid-1t Conductances .......................................................................... 92
3.4 Variation of Hybrid Parameters with IIeI, IVeEI and T ............................................................ 99
3.5 The Parameters/T ............................................................................................................... 102
Expression for I~ ..................................................................................................................................................... 104
Current Gain with Resistance Load: .................................................................................. 105
Miller's Theorem ................................................................................................................ 106
3.6
3.7
3.8
3.9
3.10
3.11
3.12
3.13
3.14
CE Short Circuit Current Gain ............................................................................................
Hybrid -1t (pi) Parameters ..................................................................................................
Measurement of IT ...................................................................... :.....................................
Variation ofHybrid-1t Parameter with Voltage, Current and Temperature ...........................
Specifications ofAmplifiers ...............................................................................................
Design of High Frequency Amplifiers ................................................................................
107
116
121
121
126
127
Objective Type Questions ............................................................................................................ 141
Essay Type Questions ................................................................................................................. 142
Answers to Objective Type Questions ........................................................................................ 143
Unit - 4
Power Amplifiers ............................................. 145
4.1
Introduction ....................................................................................................................... 146
42 Class A Power Amplifier ..................................................................................................... 148
xiii
4.3 Maximum Value of Efficiency of Class A Amplifier .............................................................
4.4 Transformer Coupled Amplifier ..........................................................................................
4.5 Transformer Coupled Audio Amplifier ................................................................................
4.6 Push Pull Amplifiers ...........................................................................................................
4.7 Complimentary Symmetry Circuits (Transformer Less Class B Power Amplifier) ................
4.8 Phase Inverters ..................................................................................................................
4.9 Class D: Operation ...........................................................................................................
4.10 Class S: Operation ...........................................................................................................
4.1 1 Heat Sinks ..........................................................................................................................
ISO
153
157
162
168
170
174
175
ISO
Objective Type Questions ....... ..................................................................................................... 183
Essay Type Questions ................................................................................................................. 185
Answers to Objective Type Questions ........................................................................................ 186
Unit - 5
Tuned Amplifiers - I ...................................................... 189
5.1
52
5.3
5.4
5.5
5.6
Introduction ....................................................................................................................... 190
Single Tuned Capacitive Coupled Amplifier ....................................................................... 191
Tapped Single Tuned Capacitance Coupled Amplifier ....................................................... 196
Single Tuned Transformer Coupled or Inductively Coupled Amplifier ............................... 200
CE Double Tuned Amplifier ................................................................................................ 204
Applications of Tuned Amplifiers ...................................................................................... 208
Objective Type Questions .............................................. '" ........................................................ '2JJJ
Essay Type Questions ............................................................................................................... 210
Answers to Objective Type Questions ...................................................................................... 211
Unit - 6
Tuned Amplifiers - II ..................................................... 21 3
6.1
62
63
6.4
6.5
6.6
Stagger Tuning .................................................................................................................. 214
Single Tuned Transistor Amplifier ...................................................................................... 214
Stability Considerations ..................................................................................................... 214
Tuned Class B and Class C Amplifiers ............................................................................... 216
WidebandAmplifiers .......................................................................................................... 221
Tuned Amplifiers .. ,............................................................................................................. 225
Objective Type Questions ......................................................................................................... 234
Essay Type Questions ............................................................................................................... 235
Answers to Objective Type Questions ...................................................................................... 236
xiv
Unit - 7
Voltage Regulators ....................................................... 237
7.1
72
7.3
7.4
Introduction ....................................................................................................................... 238
Tenninology ....................................................................................................................... 247
Basic Regulator Circuit ....................................................................................................... 248
Short Circuit Protection ............................................................................ :......................... 249
Objective Type Questions ......................................................................................................... 251
Essay Type Questions ............................................................................................................... 252
Answers to Objective Type Questions ...................................................................................... 253
Unit - 8
Switching and Ie Voltage Regulators .......................... 255
8.1
82
8.3
8.4
8.5
8.6
8.7
8.8
IC 723 Voltage Regulators and 3 Tenninal IC Regulators .................................................... 256
Current Limiting .................................................................................................................. '2fj2
Specifications of Voltage Regulator Circuits ....................................................................... 264
DC To DC Converter .......................................................................................................... 264
Switching Regulators ......................................................................................................... 264
Voltage Multipliers ............................................................................................................. 267
Uninterrupted Power Supply (UPS) .................................................................................... 286
Switched Mode Power Supplies (SMPS) ............................................................................ 290
Objective Type Questions .......................................................................................................... 295
Essay Type Questions ................................................................................................................ 296
Answers to Objective Type Questions ....................................................................................... 297
Appendices ................................................................................................................................. 299
Index .......................................................................................................................................... 332
References ................................................................................................................................. 335
SYMBOLS
gb'e
Input Conductance of BJT in C.E configuration between fictitious base terminal
B' and emitter terminal E.
}
Input impedance (resistance) of BJT in 1C:E configuration
Forward short circuit current gain in C.E configuration
Reverse voltage gain in C.E. configuration
Output admittance in C.E. configuration
Base spread resistance between base terminal B and fictitious base terminal B'.
Emitter junction capacitance
Collector junction capacitance
Transconductance or Mutual conductance
KT
T
Volt equivalent of temperature = --e
11,600
Diode constant 11 = I for G e; 11 = 2 for SI
Feedback conductance between B and collector terminal C
I
Output conductance between Collector and Emitter terminals.
Diffusion capacitance
Charge
Diffusion constant for minority carriers in Base region
constant (= 112 for abrupt junctions)
w
Base width
co
Angular frequency = 21t1
IT
Frequency at which C.E. short circuit current gain becomes unity
I~
Frequency at which h fe becomes 0.707 hfe
as the B. W of the transistor circuit.
max'
Frequency range upt0lp is referred
xvi
Cx
Incrimental capacitance in hybrid - 1t model
Incrimental resistance in hybrid - 1t model
rx
Voltage gain of I stage am~lifier circuit
Current gain of I stage amplifier circuit
AVI
All
B.W
Ie
A=J;
IH=
h.
Band width of the amplifier circuit.
Cut-off frequency
Lower cutoff frequency or Lower 3-db point or Lower half power frequency
Upper cutoff frequency or upper 3-db point or upper half power frequency
.J II
10
Mid Band Frequency 10 =
RE
CE
Emitter Resistor
Av(L.F)
Ay(H.F)
AVL = Voltage gain in the Low frequency range
AVH = Voltage gain in the High frequency range
AVM = Voltage gain in the Mid frequency range
Output Power
Input Power
Power Gain
Av(M.F)
Po
PI
Ap
~
Vy
Iy
1m
Vm
Ip _ p
Pac
PDe
TJ
n
N2
NI
VI
V2
~
Rp
Vbe
VSE
12 .
Emitter Capacitor
Phase angle
RMS value of voltage
RMS value of current
(IMax - IMm )
(V Max - V Min)
Peak to Peak value of current
A.C. Output power
DC Input power
Conversion Efficiency of the power amplifier circuit.
Transformer turns ratio (N21N I)
Number of turns of transformer Secondary winding
Number of turns of transformer Primary winding
Primary voltage of Transformer
Secondary voltage of Transformer
Resistance of Tuned Circuit
Parallel resistance associated with the tuning coil (Inductor)
A.C voltage between base and emitter leads of transistor (BJn
D.C voltage between base and emitter leads of transistor (BJT)
Small subscripts are used for a.c. quantities.
Capital subscripts are used for a d.c. quantities.
xvii
Oe
a
Effective Q factor of coil
Ru
Resistance of tapped tuned Circuit
Qo
M
Quality factor of output circuit
Kc
Critical value of the coefficient of coupling
Me
Critical value of Mutual Inductance
Rs
S
Series Resistance in Voltage Regulators
Sr
Temperature coefficient in Voltage Regulator
Ro
Output Resistance
Rz
Zener Diode Resistance
Vy
Cut in voltage of junction diode
Fractional Frequency Variation
Mutual Inductance
Stability factor
VO'(P_P)
Output ripple voltage
Vj'(P_P)
Input ripple voltage
=
"This page is Intentionally Left Blank"
Brief History of Electronics
In science, we study about the laws of nature and verification and in technology, we study the
applications of these laws to human needs.
Electronics is the science and technology of the passage of charged particles, in a gas or
vacuum or semiconductor.
Before electronic engineering came into existence, electrical engineering flourished. Electrical
engineering mainly deals with motion of electrons in metals only, where as Electronic engineering
deals with motion of charged particles (electrons and holes) in metals, semiconductors and also in
vacuum. Another difference is, in dectrical engineering, the voltages and currents are very high KV,
and Amperes, where as in electronic engineering one deals with few volts and mAo Yet another
difference is, in elect{ical engineering; the frequencies of operation are 50 Hzs/60Hzs. In electronics
it is KHzs, MHz, GHzs, (high frequency).
The beginning for Electronics was made in 1895 when H.A. Lorentz postulated the existence of
discrete charges called electrons. Two years later, JJ Thomson proved the same experimentally
in 1897.
In the same year that is in 1897, Braun built the first tube based on the motion of electrons, the
Cathode ray tube (CRT).
In 1904 Fleming invented the Vacuum diode called 'valve'.
In 1906 a semiconductor diode was fabricated but they could not succeed, in making it work.
So semiconductor technology met with premature death and vacuum tubes flourished.
xx
In 1906 it self, De Forest put a third electrode into Fleming's diode and he called it Triode. A
small change in grid voltage produces large change in plate voltage, in this device.
In 1912 Institute of Radio Engineering (IRE) was set up in USA to take care of the technical
interests of electronic engineers. Before that in 1884 Institute of Electrical Engineers was
formed and in 1963 both have merged into one association called IEEE (Institute of Electrical and
Electronic Engineers).
The first radio broadcasting station was built in 1920 in USA.
In 1930 black and white television transmission started in USA.
In 1950 Colour television broadcasting was started.
The electronics Industry can be divided into 4 categories:
Components
Communications
Transistors, ICs, R, L, C components
Radio, TV, Telephones, wireless, land line communications
Control
Computation
Industrial electronics, control systems
Computers
Vacuum Tubes ruled the electronic field till the invention of transistors. The difficuty with
vacuum tubes is with its excess generated heat. The filaments get heated to > 2000° K so that
electronic emission takes place. The filaments get burnt and tubes occupy large space. So in 1945
Solid State Physics group was formed to invent semiconductor devicl!s in Bell labs, USA.
1895: H. A. Lorentz - Postulated existence of Electrons
1897: J.J. Thomson - Proved the same
1904: Fleming - Vacuum Diode
1906: De. Forest - Triode
1920: Radio Broadcasting in USA
1930: Black and White TV USA
1947:
1947:
1950:
1959:
1959:
Shockley invented the junction transistor. (BJT)
Schokley BJT Invention
Colour Television
Integrated Circuit concept was announced by Kilby at an IRE convention.
KILBY etc. anounced ICs.
1969: LSI, IC : Large Scale Integration, with more than 1000 but < 10,000 components per
chip (integrated or joined together), device was announced.
1969: SSI 10 - 100 comp/chip. LOGIC GATES, FFs.
1970: Intel People, 9 months, chip with 1000 Transistors (4004!!p)
1971: !!p - 4 bit INTEL
1971: 4 bit Microprocessor was made by Intel group.
1975: VLSI: Very large scale integration> 10,000 components per chip. ICs were made.
1975: CHMOS - Complimentary High Metal Oxide Semiconductor ICs were announced by
Intel.
xxi
1975: MSI (Multiplenum, Address) 100 - 1000 comp/chip
1978: LSI 8 bit
~Ps,
ROM, RAM 1000 - 10,000 comp/chip
1980: VLSI
> 1,00,000 components/ser 16, 32 bit
~Ps
1981: 16 bit J.lP
> 1,00,000 components/ser 16, 32 bit
~Ps
1982 : 100,000 Transistors, 80286 Processor
> 2,00,000 components/ser 16, 32 bit
1984: CHMOS
~Ps
1985: 32 bit
~
P
> 4,50,000 components/ser 16, 32 bit
~Ps
1986: 64 bit
~
P
> 10,00,000 components/ser 16, 32 bit
~Ps
1987: MMICS
Monolithic Microwave Integrated Circuits
1989: 1860
Intel's 64 bit CPU
1990: ULSI > 500,000 Transistors ultra large scale
1992: GSI > 10,00,000 Transistors Giant scale
100, 3 million Transistors, Pentium
1998: 2 Million GateslDie
2001: 5 Million Gates / Die
2002: 1000, 150 Million Transistors.
1 Gigabit Memory Chips
Nature is more SUPERIOR
2003: 10 n.m. patterns, line width
2004: Commercial Super Compo IOTRILLION Flip Flops
2010: Neoro - Computer Using Logic Structure Based on Human Brain
There are 107 cells/cm3 in human brain
VLSI Technology Development:
3 ~ Technology
J,
0.5
~
Technology
J,
0.12
~
Technology
ASICs (Application Specific Integrated Circuits)
HYBRIDICs
BICMOS
MCMs (Multi Chip Modules)
3-D packages
xxii
Table 1
Table showing VLSI technology development predictions made in 1995.
1995·
1998
20'01
2004
2007
Lithography (11)
0.35
0.25
0.18
0.12
0.1
No. Gates/Die:
800K
2M
5M
10M
20M
DRAM
64M
256M
1G
4G
16G
SRAM
16M
64M
256M
IG
4G
Wafer Dia (mm)
200
200-400
400
400
400
Power (11 W/Die)
15
30
40
40-120
40-200
Power Supply. (V)
3.3
2.2
2.2
1.5
1.5
Frequency (MHz)
100
175
250
350
500
No. Bits/Die
UNIT - 1
Single Stage Amplifiers
In this Unit,
• Single stage amplifiers in the three configurations of C.E, C.B~ C.C,
with design aspects are given.
• Using the design formulae for Av' AI' Rt , Ro etc, the design of single
stJge amplifier circuits is to be studied.
• Single stage JFET amplifiers in C.D, C.S and C.G configurations are
also given.
• The Hybrid - 7t equivalent circuit of BJT, expressions for Transistor
conductances and capacitan~es are derived.
• Miller's theorem, definitions forip andfT are also given.
• Numerical examples, with design emphasis are given.
2
1.1
Electronic Circuit Analysis
Introduction
An electronic amplifier circuit is one, which modifies the characteristics of the input signal, when
delivered the output side. The modification in the characteristics of the input signal can be with
respect to voltage, current, power or phase. Anyone or all these characteristics power, or phase
may be changed by the amplifier circuit.
1.1.1
Classification of Amplifiers
Amplifier circuits are classified in different ways as indicated below:
Types of Classification
(a) Based on Frequency range
(b) Based on Type of coupling
(c) Based on Power delivered/conduction angle
(d) Based on Signal handled.
(a) Frequency Range
AF(Audio Freq.)
RF(Radio Freq.)
Video FreQuency
VLF (Very Low Fr~.)
LF (Low Frequency)
Medium Frequency
High Frequency
VHF (Very HIgh Freq-l
UHF (Ultra High Freq.)
. SHF (Super HIgh Freq.)
40 Hzs - 15/20 KHz
>20KHz
5 -8 MHz
10-30KHz
30-300KHz
300 - 3000 KHz
3 -30 MHz
30-300 MHz
300 -3000 MHz
3000 - 30,000 MHz
(b) Types of Coupling
1. Direct coupled
2. RC coupled
3. Transformer coupled
4. LC Tuned Amplifiers
5. Series fed.
(c) Output power delivered/conduction angle
1. Low power (tens of mW or less).
2. Medium power (hundreds of mW).
3. High power (Watts).
Class A
ClassB
ClassAB
Class C
Class D
Class S
3600
1800
180 - 3600
< USOO
Switching type.
Switching type.
Single Stage Amplifzers
3
(d) Type of signal handled
1. Large signal
2. Small signal
In addition to voltage amplification Av, current amplification AI or power amplification Ap is
expected from an amplifier circuit. The amplifier circuit must also have other characteristics like
High input impedance (Zl or Rj), Low output impedance (Zo or Ro)' Large Band Width (BW), High
signal to Noise Ratio (SIN), and large Figure of Merit (Gain BW product).
,
In order that the amplified signal is coupled to the load RL or Zu for all frequencies of the
. input signal range, so that maximum power is transferred to the load, (the condition required for
maximum power transfer is IZol = IZd or Ro = RL) coupling the output of amplifier V0 to load RL
or ZL is important. When reactive elements are used in the amplifier circuit, and due to internal
junction capacitances of the active device, the Zj and Zo of the amplifier circuit change with
frequency. As the input signal frequency varies over a wide range, and for all these signals
amplification and impedance matching have to be achieved, coupling of the output of the amplifier
to the load is important.
Since the gain Av, AI or Ap that can be obtained from a single stage amplifier circuit where only
one active device (BJT, JFET or MOSFET) is used, the amplifier circuits are cascaded to get large
gain. Multistage amplifier circuits are discussed in the next chapter.
When the frequency of the input signal is high (greater than A.F. range) due to internal junction
capacitances of the actual device, the equivalent circuit of the BJT used earlier is not valid. So
another model of BJT valid for high frequencies, proposed by Giacoletto is studied in this chapter.
1.2
Small Signal Analysis of Junction Transistor
Small Signal Analysis means, we assume that the input AC signal peak to peak to amplitude is
very small around the operating point Q as shown in Fig. 1.1. The swing of the signal always
lies in the active region, and so the output is not distorted. In the Large Signal Analysis, the
swing of the input signal is over a wide range around the operating point. The magnitude of the
input signal is very large. Because of this the operating region will extend into the cutoff region
and also saturation region.
Fig. 1.1 Output Characteristics of BJT
Electronic Circuit Analysis
4
1.3
Common Emitter Amplifier
Common Emitter Circuit is as shown in the Fig. 1.2. The DC supply, biasing resistors and coupling
capacitors are not shown since we are performing an AC analysis.
Fig. 1.2 C.E. Amplifier
Es is the input signal source and Rs is its resistance. The h-parameter equivalent for the above
circuit is as shown in Fig. 1.3.
The typical values of the h-parameter for a transistor in Common Emitter Configuration are,
hIe = 4 KO,
+- Ie
~----~--~~r-----~
E~
________
~
______
~
____r -____
~
____~______~____~
Fig. 1.3 h-parameter Equivalent Circuit
Since,
Vbe is a fraction of volt O.2V, Ib in
h· =
Ie
~A,
100
0.2V
=4KO
50xlO-6
~A
and so on.
C
E
Single Stage Amplifiers
5
h fe == Ic/Ib :: 100.
Ie is in rnA and Is in 1lA.
1:: P
hfe »
h re == 0.2 x
10-3 .
Because, it is the Reverse Voltage Gain.
V
h re =
and
be
Vee
Vee> V be ;
Input
h = -"'--re
Output
Output is »
input, because amplification takes place. Therefore hre «
hoe= 8 Il 70;:
van d hoe ==
1.
~.
Vee
1.3.1
Input Resistance of the Amplifier Circuit (Ri)
The general expression for RI in the case of Common Emitter Transistor Circuit is
hfeh re
1
h oe + -
..... (1.1)
RL
For Common Emitter Configuration,
hfeh re
1
h oe + -
..... ( 1.2)
RL
R; depends on RL. If RL is very small,
_1_ is large, therefore the denominator in the second
RL
term is large or it can be neglected.
R j == hIe
If RL increases, the second term cannot be neglected.
RI == hIe - (finite value)
Therefore, R j decreases as RL increases. If RL is very large, _1_ will be negligible compared
RL
to hoe' Therefore, RI remains constant. The graph showing R j versus RL is indicated in Fig. 1.4.
RI is not affected by RL if RL < 1 KQ and RL > 1 MQ as shown in Fig. 1.4.
Electronic Circuit Analysis
6
,,
10000
,,,
,,
,,
,,
-+---------,
6000
I
1 KO
1 MO RL
Fig. 1.4 Variation of Rj with RL
R, varies with frequency f because h-parameters will vary with frequency. hfe , hre will change
with frequency f of the input signal.
1.3.2
Output Resistance of an Amplifier Circuit (Ro)
For Common Emitter Configuration,
1
R = ---;-----,-o
hoe _ ( hreh fe )
hje + Rs
..... (1.3)
Rs is the resistance of the source. It is of the order of few hundred
Ro depends on Rs. If Rs is very small compared to hie'
Ro
=
Q.
1
- - h - h - (independent of Rs )
h -~
oe
h.
Ie
Then, Ro will be large of the order of few hundred KQ. If Rs is very large, then
Ro:::: - 1 :::: 150 KQ.
hoe
The graph is as shown in Fig. 1.5.
Ro
800KO
1.--_
1
200KO
--....,~~
Rs
Fig. 1.5 Variation of Ro with Rs
..... (1.4)
Single Stage Amplifiers
1.3.3
Current Gain (
7
AJ
..... (1.5)
If RL is very small, Ai::: hfe ::: 100. So, Current Gain is large for Common Emitter Configuration.
As RL increases, A, drops and when RL = co, Ai = O. Because, when RL = co, output current 10 or load
current IL = O. Therefore, A, = O. Variation of Ai with RL is shown in Fig. 1.6.
50 t----..,.
A
I
t
IMQ
-
RL
Fig 1.6 Variation of A, with RL
1.3.4
Voltage Gain (Av)
..... (1.6)
If RL is low, most of the output current flows through RL. As RL increases, output voltage increases
and hence Av increases. But if RL»
_1_, then the current from the current generator in the
hoe
h-parameters equivalent circuit flows through hoe and not RL.
I
Then the,
Output Voltage = hfe . lb' hoe
(R L is in parallel with hoe' So voltage across hoe= voltage across RL). Therefore, V0 remains constant
as output voltage remains constant (Fig.I.7).
3300
I
1 IUl
100 MO
- + RL
Fig. 1.7 Variation of Av with RL
Electronic Circuit Analysis
8
1.3.5 Power Gain
As RL increases, A~~decreases. As RL increases, Ay also increases.
Therefore, Power Gain which is the product of the two, Ay and AI varies as shown in Fig. 1.8.
Ap=AyAI
t
--+
~
Fig. 1.8 Variation of Ap with RL
Power Gain is maximum when RL is in the range 100 KQ - 1 MQ i.e., when RL is equal to the
output resistance of the transistor. Maximum power will be delivered, under such conditions.
Therefore, it can be summarised as, Common Emitter Transistor Amplifier Circuit
will have,
.
1.
')
3.
4.
5.
6.
Low to Moderate Input Resistance (300£2- 5KQ).
Moderately High Output Resistance (10K£2- 100KQ).
Large Current Amplification.
Large Vo:tage Amplification.
Large Power Gain.
1800 phase-shift between input and output voltages.
As the input current IB , increases, Ie increases therefore drop across Re increases and
Vo = Vee - VI drop across Rc- Therefore, there is a phase shift of 180°.
The amplifier circuit is shown in Fig. 1.9.
Fig. 1.9 CE Amplifier Circuit
Single Stage Amplifiers
9
1.4
Common Base Amplifier
The circuit diagram considering only AC is shuwn In Fig. 1.10.
NPN
Ie_
E
Fig. 1.10 CB Amplifier
h = Veb
ib
I
e Vcb=O
Veb is small fraction of a volt. Ie is in rnA. So, hib is small.
lei
h =fb I
e Vcb =0
hOb =
= -
VIc
0.99 (Typical Value)
= 7.7 x 10-8 mhos (Typical Value)
cb Ie=O
Ie will be very small because Ie = O. This current flows in between base and collector loop.
~b =
Veb
V
= 37 x 10-6
(Typical Value)
cb Ie = 0
~b
1.4.1
is small, because Veb will be very small and Veb is large.
Input Resistance (Ri )
hfb is -ve
when RL is small < 100 Kn, the second term can be neglected.
i
Ri = h lb :::: 30n.
when RL is very large, _1_ can be neglected.
RL
..... (1.7)
Electronic Circuit Analysis
10
So
[.:
R; ::: soon (Typical-"'alue)
R;=hjb+
~
is negative]
h h
fb rb
hob
The variation of R. with RL is shown in Fig. 1.11. R j varies from 20n to soon.
- -- - - - - - - - R
i
-~.-
•
200
Fig. 1.11 Variation of Rj with RL
1.4.2
Output
Resistance\~Ro)
R~ = ---h-r-b-h--
..... (1.8)
fb
h b - ----'=----=-
o
hib +Rs
If Rs is small,
But ~ is negative.
Ro = ---h-rb-h-fb
hob + --'-'~""hib
This will be sufficiently large, of the order of300 Kn. Therefore, value of hob is small. As ~
Ro = h1
also increases. [This will be much larger because, in the previous case, in the
ob
denominator, some quantity is subtracted from hob']
increases,
11
Single Stage Amplifiers
~The variation of Ro with Rs is shown in Fig. 1.12.
1 mn
100kQ~--
---+- Rs
Fig. 1.12 Variation of Ro with Rs
1.4.3
Current Gain (Ai)
-htb
A
= ----'=--
I
..... ( 1.9 )
l+hobRL
Ai is < I. Because h fe < I. As RL increases, AI decreases. Ai is negative due to hfb. The
variation of Ai with RL is shown in Fig. 1.13.
0.98
A
I
t
--+RL
Fig. 1.13 Variation of Ai with RL
1.4.4
Voltage Gain (Av)
-htbRL
=-----~~~---
A
........ ( 1.10 )
h ib + R L (hibhob - htbhrb)
v
As RL increases, Ay also increases. If RL tends to zero, Ay also tends to zero. CAy
as RL~ 0). The variation of Voltage Gain Ay with RL is shown in Fig. 1.14.
A
v
t
--+ RL
Fig. 1.14 Variation of ~ with RL
~
0,
12
1.4.5
ElectroJlJc Circuit Analysis
Power Gain (Ap)
Power Gain
Av increases as RL increases. But AI decreases as RL increases. Therefore, Power Gain,
which is product of both, varies with RL as shown in Fig. 1.15.
=
.;
"~
~
40
.c.
Ap
i
---+RL
Fig 1.15 Variation of Ap with RL
The characteristics of Common Base Amplifier with typical values are as given below.
l. Low Input Resistance (few 100 il).
2. High Output Resistance (Mil).
3. Current Amplification A, < 1.
4. High Voltage Amplification and No Phase Inversion
5. Moderate Power Gain (30). ": Ai < 1.
1.5
Common Collector Amplifier
The simplified circuit diagram for AC of a transistor (BJT) in Common Collector Configuration is as
shown in Fig. 1.16 (without biasing resistors).
B
Fig. 1.16 CC Amplifier
Single Stage Amplifiers
13
The h-parameter equivalent circuit of transistor in Common Collector Configuration is shown
in Fig. 1.1 7,
I
B
e
~----------~--------E
h
ES
Ie
h
oe
'V
V
ee
c~--------~~----~------~------~------~----c
Fig. 1.17 h-parameter Equivalent Circuit
Vbcl
hI(: = -1-
hoc
I
VIe
=
=
2,780 Q (Typical Value)
=
7.7 x 10- 6 mhos (Typical Value)
vec=o
b
ec Ib=O
hfc = - IIe I
b
= 100 (Typical value)
vce=o
.: Ie» lb'
hfe is negative because, IE and IB are in opposite direction.
Vbe = Vee ( Typical Value)
Because,
IB = 0, E - B junction is not forward baised.
V EB
=
O.
For other circuit viz., Common Base and Common Emitter, hr is much less than 1.
For Common Collector Configuration, hre :::: 1.
The graphs (variation with Rc) are similar to Common Base Configuration.
Characteristics
1. High Input Resistance:::: 3 Kfl (R,J
2. Low Output Resistance 30 n. (RoJ
3. Good Current Amplification Ai » 1
4. Av:S 1
5. Lowest Power Gain of all the configurations.
Since, Ay is < 1, the output voltage (Emitter Voltage) follows the input signal variation. Hence
it is also known as Emitter Follower. The graphs of variation with RL and Its are similar to Common
Base amplifier.
Electronic Circuit Analysis
14
Example : 1.1
For the circuit shown in Fig.I.18 estimate AI' A y , Ri and Ro using reasonable approximations. The
h-parameters for the transistor are given as
hfe = 100 hie = 2000
n
hre is negligible and hqe = 10-5 mhos (U) .
Ib = 100 J.LA..
Fig. 1.18 CE Amplifier Circuit
Solution:
At the test frequency capacitive reactances can be neglected. Vcc point is at ground because the AC
potenth:~l at Vcc = o. So it is at ground. R) is connected between base and ground forAC. Therefore,
R) II R2• R4 is connected between collector and ground. So R4 is in parallel with IIhoe in the output.
The A.C. equivalent circuit in terms of h-parameters of the transistor is shown in Fig. l.l9.
-+Ib
\rr-.....,---i'---. B
Fig. 1.19 Equivalent Circuit
The voltage source ~e Vee is not shown since, hre is negligible. At the test frequency of the
input signal ,the capacitors C) and C2 can be regarded as short circuits. So they are not shown in the
AC equivalent circuit. The emitter is at ground potential. Because Xc is also negligible, all the
AC passes through C3 • Therefore, emitter is at ground potential and {his circuit is in Common
Emitter Configuration.
Single Stage Amplifiers
1.5.1
15
Input Resistance
(~)
RI input resistance looking into the base is hIe only
re
htle h I
The expression for Ri of the transistor alone = hie -
I
.
[ h oe + RL
RL is very small and h re is negligible. Therefore, the second term can be neglected. So ~ of the
transistor alone is hIe. Now Ri of the entire amplifier circuit, considering the bias resistors is,
RI = hIe
II
RI
II R2
RIR2
100x22
---=
= 18 KQ
RI +R2
100 + 22
18x2
RI = 18 + 2 = 1.8 KQ
..
1.5.2
Output Resistance (Ro)
1
Ro - --,::-------:;-
..... (7.11)
1
- h - [ --"""hrehfe
oe
hie + Rs
Because, h re is negligible, Ro of the transistor alone in terms of h-parameters of the
1
transistor = - h . Now
oe
Ro of the entire amplifier circuit is,
R411 RL) = (2.1 x 10+3)
[ _1_11
hoe
=
1.5.3
2KQ.
I
II
(100 KQ)
II
(I kQ)
(I kQ) = 0.67 kQ
Current Gain (Ai)
To determine Ai the direct formula for Ai in transistor in Common Emitter Configuration is, 1
~hfeR
+ oe L
.
But this cannot be used because the input current Ii gets divided into II and lb. There is some current
flowing through the parallel configuration ofRI and R 2. So the above formula cannot be used.
V be = lb· hie
V be = 10-4 x (2000) = 0.2Y. (This is AC Voltage not DC)
Voltage across RI R2 parallel configuration is also V be.
..
Current
II =
V be
50xl0
0.2
3 = 50kA
u
= 4 IlA.
Electronic Circuit Analysis
16
Therefore) total input current,
Ii = I, + Ib = 4 + 100 = 104 IlA.
10 is the current through the 1Kn load.
1
- h = 100 Kn is very large compared with R4 and RL. Therefore, all the current on the
oe
output side, hfe Ib gets divided between R4 and RL only.
Therefore;current through RL is 10 ,
= 6.78 rnA.
Therefore ,current amplification,
I
A=-..2...
I
I.
I
6.78xI0-3
- - - - ; - =65.
104 x 10-6
Vo =-10' RL
= (- 6.78 x 103) x (10 3)
=-6.78V
Because, the direction of 10 is taken as entering into the circuit. But actually 10 flows down,
because Vo is measured with respect to ground.
= -
33.9
Negative sign indicates that there is phase shift of 1800 between input and output voltages,
i.e. as base voltage goes more positive, (it is NPN transistor), the collector voltage goes
more negative.
Single Stage Amplifiers
17
Example : 1.2
For the circuit shown, in Fig. (1.20), estimate Av and
~.
_1_ is large compared with the load seen
hoe
by the transistor. All capacitors have negligible reactance at the test frequency.
hie
Solution:
=
1KO, hfe = 99
~e is negligible.
Fig. 1.20 AC Amplifier Circuit (Ex: 1.2)
The same circuit can be redrawn as,
R,
20kO
50kO
+
Fig. 1.21 Redrawn Circuit of AC Amplifier
In the second circuit also, R4 is between collector and positive of Vcc. R\ is between +Vcc and
base. Hence both the circuits are identical. Circuit in Fig. 1.20 is same as circuit in Fig. 1.21. In the
AC equivalent circuit, the direct current source should be shorted to ground. Therefore, R4 is between
collector and ground and R\ is between base and ground. Therefore, R4 is in parallel with R7 and R\
is in parallel with k3 ( Fig. 1.21).
Electronic Circuit Analysis
18
B
B
r
RzIIR3
h.
Ie
J. hr.l h
V.I
E
~Ic
C
C
r
V
0
1
131
E
E
Fig. 1.22 h - Parameter equivalent circuit
60x30 1800
R211 R3 = 60+30 =~ = 20Kn.
5x20
R411 R7 = RL = 5+20 = 4Kn.
Therefore, the circuit reduces to, (as shown in Fig. 1.23).
20Kn
h~
V
o
V.
I
E
Fig. 1.23 Simplified circuit of Fig. 1.22
VI
1=b
hie
Vo
hfeRL
Ay= Vi = hie
Ay=-400
( .: hre is negligible)
-99 (14.28xl03 )
3
10
Single Stage Amplifiers
~
19
is the parallel combination of 20KQ and hie"
20xlKQ
- - - = 950Q
20+1
Example : 1.3
4..-
Given a single stage transistor amplifier with
parameter as hic = 1.1 KQ, hrc = 1, hfc = -51,
hoc = 2S /lA/v. Calculate AI' Av, Aw Ri, and Ro for the Common <:ollector Configuration, with
Rs= RL = 10K.
Solution:
\
-h fc
51
1'1.1 = {1+h RJ = 1+25x1O-6 xl0 4 =40.8
oc
RI = hlc + hrc AI RL = 1.1 x 103 + i x 40 . 8 x 104 = 409.1 KQ
AI·R L
Av =
R.
1
40.8x 10 4
409.lx10 3 = 0.998
0.998 x 409.1
= 0.974
419.1
2Sxl0-6 +
Ro = 217Q
Example : 1.4
For any transistor amplifier prove that
R =
I
h·
1
l-hrA v
Solution:
But
-h f
A = ---'-I
l+ho.R L
~ = hi + hr AI RL
Slxl
(1.1 + 10)103
4.625 x 10-3
Electronic Circuit Analysis
20
Av=
AI·R L
R.
I
Av·Ri
A
I
Substituting this value ofRL in equation (1)
RL =
Ri = hi +
hr .A I .A V .R i
A
= hi + hr· Av· ~
I
Example : 1.5
For a Common Emitter Configuration, what is the maximum value of RL for which R\ differs by no1
more than 10% of its value at ~2 = 0 ?
hfe = 50
hie = 1100n;
h re = 250 x 10--4;
hoe = 25Jl A/v
Solution:
Expression for
~
is,
hfe·h re
h
1
oe
+RL
If R2 = 0, RI = hIe. The value of RL for which Ri = 0.9 hie is found from the expression,
_
0.9 hie - hie -
or
hfe .h re
h
oe
+_1_
RL
h fe .h re
--"~- =
O.lh ie
or
hfe·hre
1
h +oe RL
~=
RL =
1
hoe + RL
0.1 hie
0.1 x 1100
4
hfeh re -0. lh 6e hie - 50x2.5xlO- -O.lxllOOx25xl0-6
11.3Kn
-
Single Stage Amplifiers
21
1.6
JFET Amplifiers
1.6.1
Common Drain Amplifier
The circuit is shown in Fig.I.24.
.-----~----o
+ V DD
Fig. 1.24 Common drain amplifier circuit
+
Fig. 1.25 Equivalent circuit
KVl
Vin - Vgs = Vo
Vo = Rs' . gm [V in - Vol
where
Rs'
=
Rs
II rd
Vo = Rs'gm Vin - Rs'gmVo
Vo
x
Vo
Vm
1.6.2
[I + Rs'gml
=
Rs'gm Vin
Rs' gm
l+Rs' gm
Common Gate Amplifier
Small signal equivalent of Fig. 1.26.
+VDD
-Vss
Fig. 1.26 Common Gate Amplifier
12
Electronic Circuit Analysis
S -r-----r----I
D
+
+
G
Fig. 1.27 (a) Equivalent circuit
Vo=-iDRD
-i
D
=
0-
om
V + Vds
gs
rd
JFET amplifier has very small gate leakage current. G - S junction is a reverse biased P - N
junction. In the ideal case, IG = O. This is equivalent to saying, ID = Is ( .: IG = 0).
JFET amplifier has high input impedance, therefore the input side i.e, G - S junction is reverse
biased. [For a bipolar transistor amplifier circuit, the input side i.e., E - B junction is forward. So it
has less resistance].
.....
23
Single Stage Amplifiers
The price paid for high input resistance is less control over output current. That is, a JFET
takes larger changes in input voltage to produce changes in output current, therefore a JFET amplifier
has much less voltage gain than a bipolar amplifier.
Equation
10
~ loss [1- ~s
r
This is a square law. This is another name for parabolic shape. So JFET is often called as a
square law device.
Example : 1.6
What is the DC input resistance of a JFET which has
(pA) at 20 V?
IGSS
(gate leakage current) = 5
picoampere~
Solution:
RGS =
Input resistance,
20V
----c-:- =
5xlO- 12
1.6.3
4
x
10 12 n
Common Source (CS) Amplifier
When a small AC signal is coupled into the gate, it produces variations in gate - source voltage. This
produces a sinusoidal drain current. Since AC flows through the drain resistor, we get an amplified
AC voltage at the output.
IMn
-VG
Fig. 1.27 (b) Common Source Amplifier
This is FET equivalent of common emitter transistor amplifier in BJT. As common emitter
configuration is widely used, C.S. configuration is also widely used in the case of FETs. As in the
case of common emitter configuration, C.S configuration also introduces a phase-shift.-O! 180°.
Vos = Voo - 10 . RL (DC values)
Electronic Circuit Analysis
24
v
Suppose AC input of positive voltage (positive half cycle
t
8 . . . ." )
is applied. Gate is
p - type. So reverse bias of G - S junction is reduced. So ID increases, because VDS decreases.
Therefore, with respect to AC, initially AC is zero, now the AC signal (output) is negative, therefore,
VD decrease. So for positive input AC, the AC output i s negative. Hence it introduces a phase-shift.
Low frequency equivalent circuit (e)
Capacitance and DC voltages are shorted. FET is replaced by its small signal model so the
equivalent circuit is as shown in Fig. 1.27 (c).
G .....-----.
s.,.t----L..-----.L..---'--....L--~s
Fig. 1.27 (c) C.S amplifier
IL
=
AC drain current
rd
=
drain resistance
Vo
=
Id RL
V.
Voltage gain = Av = ~
v.-
I
Single Stage Amplifiers
25
But
RL
rd»
+ RL ::::: rd
rd
IAv::::: -gm~:d.RL
IAv ::::: I
gm . RL
1.6.4
Phase Inversion
An increase in gate - source voltage produces more drain current, which means that drain voltage
is decreasing because drop across Ro increases.
V 00 -
10
So increasing input voltage
there is phase inversion.
1.6.5
Ro
= Vos decreases.
f'\ produces \..J decreasing output voltage. Therefore
Voltage Gain
Vo =
-
gm Vgs Ro
(phase inversion for negative sign)
Vm=Vgs
A
1.6.6
Vo
v
=-
V.
I
Distortion
The transconductance curve of a JFET is non-linear. It follows square law. Because of this,
JFET distorts large signals. This is known as square law distortion.
",,-VGS
Fig. 1.28 Transfer Characteristic of JFET
1.6.7
Swamping Resistor
The resistor that is connected in series with the source resistance Rs is called Swamping resistor. The
source resistance Rs is bypassed by Cs' the source bypass capacitor. (Similarly to RE in BJT circuit)
26
Electronic Circuit Analysis
VI
0---11---1---+-+1
Fig. 1.29 Circuit with Swamping resistor
VI
Fig. 1.30 Equivalent circuit
If rs is not there, source is at ground potential. But when rs is connected, AC current is passing,
through this, because source is not at ground potential. Thus local feedback will help in reaching the
non-linearity of the square law transconductance curve of JFET. So distortion will be reduced.
The voltage gain will be (- RD / r). So the effect of swamping resistor is,
1. It reduces distortion.
2. It reduces voltage gain.
Vgs+~Vgsrs-Vin=O
Yin = (1 + gm rs) Vgs
VO=-~ VgsRD
Vo
=?
Yin
.
Av= rs
If ~ is large,
+~m)
IAv ~ -RrsD I
"
Single Stage Amplifiers
1. 7
27
Common Drain (CD) Amplifier
It is similar to CC amplifier or emitter follower. So it is also called source follower.
1. Av < 1.
2. No phase change.
3. Less distortion than a common source amplifier, because the source resistor is not
bypassed.
VO=gm VgsRs
Vgs + gm Vgs Rs - VIn = 0
VIn = (l + gm Rs) Vgs
Vo
Vm
1.8
gmRs
l+gm R s
Common Gate Amplifier (CG)
The input IZI of CG amplifier is low. Therefore, its application is very less.
1
= 500
gm
=-
Z
In
Vo
=
gm Vgs Ro
VIn =Vgs
n
Electronic Circuit Analysis
28
Example : 1.7
For the JFET amplifier circuit shown in Fig. 1.31, if gm = 2500 ~(j, and Vm = 5 mV, what is the value
ofVo?
r - - - - , - - - + 30 V
1 MO
V,
0---1
1---+--1-+1
""":""'.f---i r - - , - - - O Vo
Ca
1 MO
3 KO
Fig. 1.31 Circuit diagram of Ex : 1.7
Solution:
gm
1
--=4000
2500 fl
Open circuit output voltage, that is without considering RL , = (0.949) (5mV) = 4.75 mY.
Output impedance is, Zo = Rs
II -
1
gm
= 7500 0
II 400
0
=3800
This is the AC equivalent circuit. An AC source of 4.75 mV is in series drawn circuit with an
output impedance of 380 O.
5mV::1
I
3800
500n
Fig. 1.32 AC Equivalent circuit
Therefore, AC voltage across the load resistor is,
3000
Vo = 3380 x 4.75 mV
= 4.22 mV
Single Stage Amplifiers
29
1.9 Gain - Bandwidth (B.W) Product
This is a measure to denote the performance of an amplifier circuit. Gain - B. W product is also
referred as Figure of Merit of an amplifier. Any amplifier circuit must have large gain and large
bandwidth. For certain amplifier circuits, the midband gain Am may_be large, but not Band width or
Vice - Versa. Different amplifier circuits can be compared with thus parameter. The expression for
Gain - B. W product of BJT.
where C s is shunt capacitor. Atf= fT' AI
=
1. So value of gain - B.W product isfT itself.
for high frequency amplifiers, the approximate expression for
'fr'
is,
Electronic Circuit Analysis
30
1. The units of h-paramt,cers are ............................ ..
2. h-parameters are named as hybrid parameters because .............................. .
3. The general equations governing h-parameters are
VI
............................ ..
12
............................ ..
4. The parameter h
re
is defined as h
re
= ............................. .
5. h-parameters are valid in the .............................. frequency range.
6. Typical values ofh-parameters in Common Emitter Configuration are
7. The units of the parameter hare ............................. .
rc
8. Conversion Efficiency of an amplifier circuit is ............................. .
9. Expression for current gain AI in terms of h and hre are AI = ........................ .
fe
10. In Common Collector Configuration, the values of hrc- ............................. .
11. In the case of transistor in Common Emitter Configuration, as l\. increases, R, ................. ..
12. Current Gain AI ofBJT in Common Emitter Configuration is high when RL is ................... .
13. Power Gain of Common Emitter Transistor amplifier is ................... .
14. Current Gain AI in Common Base Configuration is ................... .
15. Among the three transistor amplifier configurations, large output resistance is in ................... .
configuration.
16. Highest current gain, under identical conditions is obtained in .................... transistor amplifier configuraiton.
17. C.C Configuration is also known as .................... circuit.
Single Stage Amplifiers
31
1. Write the general equations in terms ofh-parameters for a BJT in Common Base Amplifiers
configuration and define the h-parameters.
2. Convert the h-parameters in Common Base Configuration to Common Emitter Configuration,
deriving the necessary equations.
3. Compare the transistor ( BJT) amplifiers circuits in the three configurations with the help of
h-parameters values.
4. Draw the h-parameter equivalent circuits for Transistor amplifiers in the three configurations.
5. With the help of necessary equations, discuss the variations of Av' AI' R, Ro' Ap with ~ and
l\ in Common Emitter Configuration.
6. Discuss the Transistor Amplifier characteristics in Common Base Configuration and their
variation with ~ and l\ with the help of equations.
7. Compare the characteristics of Transistor Amplifiers in the three configurations.
Electronic Circuit Analysis
32
============::::::::::j. ns" ers to Ohjecti\ e Questions.~=========
1.
n, mhos and constants
2.
the units of different parameters are not the same
3.
hllIl+h12V2
h21 I 1 + h22 V 2
4.
5.
Audio
6.
h. = 1Kn, h = 1.5 x 10 ,
-4
Ie
re
hoe = 6Jlmhos, hfie = 200
7.
8.
No units (constant)
AC Signal Power Delivered to the Load xlOO
DC Input Power
9.
10.
11.
Decreases
12.
Low
13.
Large
14.
<1
15.
Common Base configuration
16.
Common Collector
Configuration
17.
Voltatge follower/buffer
UNIT - 2
Multistage Amplifiers
In this Unit,
• Cascading of single stage amplifiers is discussed.
• Expressions for overall voltage gain are derived.
• Lower cut-off frequency, upper cut-off frequency, when n-stages are
cascaded are given.
• Phase response of an amplifier, decibel voltage gain, stiff coupling terms
are explained.
• Other types of transistor circuits, Darlington pair circuits, Boot strapped
sweep circuit, Cascode amplifiers are also discussed.
• Numerical examples are also given.
34
2.1
Electronic Circuit Analysis
Multistage Amplifiers Methods of Inter Stage Coupling
If the amplification obtained from a single stage amplifiers is not sufficient, two or more such amplifiers
are connected in Cascade or Series i.e., the output of the first stage will be the input to the second
stage. This voltage is further amplified by the second stage and so we get large amplification or large
output voltage compared to the input. In the multistage amplifiers, the output of the first stage should
be coupled to the input of the second stage and so on : Depending upon the type of coupling, the
multistage amplifiers are classified as :
1. Resistance and Capacitance Coupled Amplifiers (RC Coupled)
2. Transformer Coupled Amplifiers
3. Direct Coupled DC Amplifiers
4. Tuned Circuit Amplifiers.
2.1.1
Resistance and Capacitance Coupled Amplifiers (RC Coupled)
This type of amplifier is very widely used. It is least expensive and has good frequency response.
In the multistage resistive capacitor coupled amplifiers, the output of the first stage is coupled to the
next through coupling capacitor and RL. In two stage Resistor Capacitor coupled amplifiers, there is
no separate RL between collector and ground, but Reo the resistance between collector and Vcc (Rc)
itself acts as RL in the AC equivalent circuit.
2.1.2
Transformer Coupled Amplifiers
Here the output of the amplifier is coupled to the next stage or to the load through a transformer. With
this overall circuit gain will be increased (.:
J
N2 =
V2 and also impedance matching can be achieved.
NI
VI
But such transformer coupled amplifiers will not have broadfrequency response i.e., (f2 -f,) is small
since inductance of the transformer windings will be large. So Transformer coupling is done for
power amplifier circuits, where impedance matching is critical criterion for maximum power to be
delivered to the load.
2.1.3
Direct Coupled (DC) Amplifiers
Here DC stands for direct coupled and not (direct current). In this type, there is no reactive element.
Lor C used to couple the output of one stage to the other. The AC output from the collector of one
stage is directly given to the base of the second stage transistor directly. So type of amplifiers are
used for large amplification of DC and using low frequency signals. Resistor Capacitor coupled
amplifiers can not be used for amplifications of DC or low frequency signals since Xc the capacitive
reactance of the coupling capacitor will be very large or open circuit for DC
(Xc = 112 rtfc. Iff= 0 or low, then Xc ~ 00.)
2.1.4
Tuned Circuit Amplifiers
In this type there will be one RC or LC tuned circuit between collector and Vceo in the place of Re.
These amplifiers will amplify signals of only fixed frequency..fo which is equal to the resonance
frequency of the tuned circuit LC. These are also used to amplify signals of a narrow band of
frequencies centerd around the tuned frequency 10.
Multistage Amplifiers
35
Bandwidth of Amplifiers
2.1.5
The gain provided by an amplifier circuit is not the same for all frequencies because the reactance of
the elements connected in the circuit and the device reactance value depend upon the frequency.
Bandwidth of an amplifier is the frequency range over which the amplifier stage gain is reasonably
constant within ± 3 db, or O. 707 ofA V Max Value.
Based upon the B. W. of the amplifiers, they can be classified as :
1. Narrow hand amplifiers: Amplification is restricted to a narrow band offrequencies around
a centre frequency.
There are essentially tuned amplifiers.
2. Untuned amplifiers : These will have large bandwidth. Amplification is desired over a
considerable range of frequency spectrum.
Untuned amplifiers are further classified w.r.t bandwidth.
I.
2.
3.
4.
2.1.5.1
DC amplifiers (Direct Coupled)
Audio frequency amplifiers (AF)
Broad band amplifier
Video amplifier
DC to few KHz
20 Hz to 20 KHz
DC to few MHz
100 Hz to few MHz
Distortion in Amplifiers
If the input signal is a sine wave the output should also be a true sine wave. But in all the cases it may
not be so, which we characterize as distortion. Distortion can be due to the nonlinear characteristic
of the device, due to operating point not being chosen properly, due to large signal swing of the input
from the operating point or due to the reactive elements Land C in the circuit. Distortion is
classified as :
(a)Amplitude distortion: This is also called non linear distortion or harmonic distortion. This
type of distortion occurs in large signal amplifiers or power amplifiers. It is due to the
nonlinearity of the characteristic of the device. This is due to the presence of new frequency
signals which are not present in the input. If the input signal is of 10KHz the output signal
should also be 10KHz signal. But some harmonic terms will also be present. Hence the
amplitude of the signal (rms value) will be different Vo = Ay Vi. But it will be V<>.
(b) Frequency distortion: The amplification will not be the same for all frequencies. This is
due to reactive component in the circuit.
(c) Phase - shift delay distortion: There will be phase shift between the input and the output
and this phase shift will not be the same for all frequency signals. It also varies with the
frequency of the input signa\.
In the output signal, all these distortions may be present or anyone may be present because
of which the amplifier response will not be good.
Example : 2.1
The transistor parameters are given as,
he
hre = 6 x 10-4
hoe -- 25 r-IIAN
hIe = 2 K
uti = 50
hic = 2 K
hfc = - 51
h rc = 1
hoc = 25 IlAN
Find the individual as well as overall voltage gains and current gains.
36
Electronic Circuit Analysis
+
+
v
s
RI2
Fig. 2.1 (a) Circuit for Ex : 2.1
Solution:
It is advantageous to start the analysis with the last stage. Compute toe current gain first, then
the input impedance and voltage gain.
II stage:
For the second stage RL
=
Re2
(negative (-) sign is there because for NPN
transistor, Ie is negative (-) since it is leaving
the transistor)
51
1+ 25 x 10-6 x 5 x 103
= 45.3
Input impedance
Rj2 = hjc + h rc AJ2 Re2
= 2 + 45.3 x 5K = 228.5 K.
:. Input Z of the C.C. stage is very high.
Voltage gain of the second stage
Vo
Re2
Av =-=A 2
V2
12 Ri2
Vo = Ie2 Re2 ;
V2 = Input voltage for the second stage
45.3x5
228.5
=
0.99
=
Rj . Ij
2
Multistage Amplifiers
I stage:
37
For the first stage, the net load resistance in the parallel combination ofRcl and Ri2 or
RCJ .R 12
5 x 228.5
R12
R
R
= 4.9 KQ
CJ + 12
233.5
-50
Hence
1+ 25 x 10-6 x 4.9 X 103
= -
44.5
The input impedance of the first stage will also be the input Z of the two stages since input Z
of the second stage is also considered in determining the value of RLJ" Ril depends on RLJ"
R II = h.Ie + h re AI I RL I
(from the standard formula)
= 2 - 6 x 10-4 x 44.5 x 4.9
..
Ril
=
1.87 KQ.
Voltage gain of the first stage is,
- 44.5x4.9 '. = _ 116.6
1.87
In the actual circuit, the current gets branched into lei and Ib2 which depend upon the values
ofRc and RI
I
2
(- Ib2 + lei) Rei
+ Ib2 (Ri2 + Rei)
=
=
Ib2 . Ri2
lei . Rei
Fig. 2.1 (b) Current branching
38
Electronic Circuit Analysis
4S.3x(-44.S)xS
228.S+S
AI =-43.2
Vo
Ay
1
Av
2.1.6
Vo V
2
= y=y- .y=A
V2 ·AVI
=
2
I
0.99 x (-116.6)
= -
lIS.
R - C Coupled Amplifier Circuit (Single Stage)
c~
1
Fig. 2.2 RC Coupled amplifier circuit
RL is the load resistor that develops the output voltage from the transistor. C2 is used to couple
the AC component of the output to RL"
In the self bias circuit, RE the emitter resistors is connected between emitter and ground. Btit
through RE, a negative feedback path is there. So to prevent A.C. negative feedback, if a capacitor is
connected in parallel with RE , and it is chosen such that XE provides least resistance path compared
to RE, AC signal passes through CE and not through RE• Therefore, there will not be any negative
feedback for AC signals.
RE
XE is chosen such that X E ::; 10
and XE is chosen at the frequency fi. For frequencies higher thanfi, XE any way will be less.
Multistage Amplifrers
39
The equivalent circuit for the above transistor configuration is,
hIe
C
Fig. 2.3 Equivalent circuit
Suppose the value of Cc is very large. Then at the low and medium frequencies, the impedance
is negligible. So the equivalent circuit is as shown. We shall consider the effect of CE later.
From the equivalent circuit,
Vo = - hfe · i b· RL
negative sign is used since it is NPN transistor. The collector current is flowing into the transistor.
Input current,
Ib =
Rs + R j
Vi
For a transistor amplifier circuit in common emitter configuration,
R j = hje + (1 + h fe ) ZE'
Where
ZE = RE in parallel with C E =
1+ jroC E RE
Vs
Vo
A=y
~
S
Ay at low frequencies, (LF)
Ay(LF)
Vo
=
V.
S
When ro is large, (l+hfe)R E can be neglected. So in the Mid Frequency range, (M.F.),
1+ jro C E RE
-hfeRL
Av (M.F) = R
h
s + ie
In this expression, there is no for ro term. Hence in the mid frequency range, Ay is independent
off or the gain remains constant, irrespective of change in frequency.
40
2.1. 7
Electronic Circuit Analysis
Effect of Coupling Capacitor on Low Frequency Response
Suppose that the value of CE is such that its effect on the frequency response can be neglected and
RI
(b)
Fig. 2.4 Effect of coupling capacitors
the value of Xc at low frequencies is such that it is not a simple short circuit for A.C. signals, so that
its effect has to be considered.
The effect of CE is, voltage drop across Cc will reduce Vi with a corresponding drop in Vo.
The frequency at which the gain drops by a factor of
where
and
1
fi' is the lower 3 db frequency.
R/ = RI II R2 II Ri
Ri = hie' for an ideal capacitor CEo
These expressions are valid when emitter is bypassed. In AC equivalent circuit, emitter is
assumed to be at GND potential. Therefore CE has no effect.
Large values of capacitors are required (1 OJ.lF, 5 J.lF etc.) in transistor amplifiers for coupling
and bypass purposes for good low frequency response. Since these values are large, only electrolytic
capacitors are used. Capacitors of such large values are not available in other types of capacitors.
These capacitors are bulky. So audio amplifier circuits, Integrated Circuits (I.C) often have large
capacitors connected externally to the (I.C) itself to provide the required low frequency response.
Example : 2..2
lIn the Resistance Capacitance (RC) coupled amplifier, Av = 50, fl = 50 Hz and
m
100 K Hz. Find the values of frequencies at which the gain reduces to 40 on either side of
midband region.
fi
=
41
Multistage Amplifiers
Solution:
AYMF
AVH~ 1+ j(~)
Phase shift angle
~ ~ 180' - tan-{~)
r
AYMF
AVLF~ ~I+(;
At the frequency f, the gain is Ay L.F
AyL.F = 40,
=
40.
AYM .. F = 50,
I
40
fi
= 50 Hzs,f= ?
:. f= 66.66 Hz.
50 ~ ~I+(~r
A YM .F
A ~ ~I+(~r
VH F
.
At frequency f,
f= ?
A YH .F =40.
40
50
fi =
100 KHz.
f= 75 KHz.
1+[
f
100xl03
]2'
Electronic Circuit Analysis
42
Example : 2.3
An amplifier of 40db gain has both its input and output load resistances equal to 600ft If the
amplifier input power is - 30db, what is the output power and voltage? Assume that reference
power level used is 1m W in a 600 n resistance.
Solution:
Since it is logarithmic scale, for dbs,
Po (db) = (Pj)db + (gain db);
Po = P j x Ap Output power (Po) = Input Power (P)
x
Power Gain (Ap)
Po = ?P j = -30db, gain = 40db
Po = -30 + 40 = IOdb.
Reference power level = 1m W.
(1:~)= lOdb
10 log =
Po
10 log 0.001 = 10
Po
log 0.001 = 1
Po
--=10
0.001
log 10 = 1 .
Po = 10 x 0.001
=
10mW
(With respect to, 1m W, the input and output powers are compared and expressed in db)
Po
10 log ImW = lOdb)
RMS output voltage is,
v,2
o
P =o
R
or
Vo =
=
~Po·R =~10xl0-3 x600
16
= 2.45 V
43
Multistage Amplifiers
Example : 2.4
An amplifier has ~ = O.SKQ and Ro = O.OSKQ. The amplifier gives an output voltage of 1V peak for
an input voltage of 1mV peak. Find Av, Ai' Ap in db.
Solution:
V0 (peak)
=
I V.
VO(peak)
Vo
IV
.fi
(rms) =
Yo (rms)=
.fi
~ 20 log [~~ )
Pv
=
2~ log (.~Y)
~i (peak)
= 60 db
=
Yi (Peak) / Ri
ImY/SOO
=
2
=
10-6 A.
x
10 (peak) = YO (Peak) / RO
1ISO
=
0.02 A.
10 (Peak) =
0.02 )
.. Current Gain (Ai) = 20 log ( 2xlO-6 = 80 dl'
y2
Rj
p. =
_I
I
=
,
Y02
Po =
Power gain,
Ap
R
10-9 W.
o
= 1O-2 W.
~ 10 log [:~ ) ~ 70 db
2.2
n - Stage Cascaded Amplifier
2.2.1
Overall Voltage Gain
The resultant voltage gain is given by the product of the individual voltage gains of each stage.
Y
AVI
=
V2I
=A 1 L. 9 1
44
Electronic Circuit Analysis
Al == Magnitude of the voltage gain of the first stage and 9 1 is the phase angle between output
and input voltage of this stage
Vo == V2
V3
VI
VI' V2
A..;
.
~
V4
V3
Vo
Vn_I ' Vn
== A YI A Y2 .... A yn
== AYI L9 I" AY2 L92 . AY3 L93
= AYI AY2 ..... AYn
= Ay
2.2.2
.•..... AYn
L9 1 + L92 + L93
L9 n
L9 n
.•.••..
[Since Ay = AYI AY2 AY3 ... AYn and 9 = 9 1 + 92 + ... + 9n]
L9.
Current Gain
10
A
=-
Ibl
I
It is the ratio of the output current 10 of the last stage to the input current or base current Ib
l
of the first stage.
A
~
=
Ibl
I
=
-len
Ibl
Where len is equal to the collector current of the nth stage transistor.
~=i2
Ib}
~
I bl ' II
In_1
= A'II A'12 ........ A' In
AII
A\
=
n
=
The base to collector current gain of the first stage
Base collector current gain of the nth stage
1 =1
1\
c, '
C,
B,
Q,
E,
B2
R
E,
Q2
c,
E2
I stage
1 =1
12
cn
C2
R
C2
R
II stage
Fig. 2.5 Multistage amplifiers
2.2.3
Power Gain
A
=
P
But
Output power
Input power
-....!..-----!---
nth stage
n
Multistage Amplifiers
45
A
But
v
=
Output Voltage
Input Voltage
--=-----=-
I/Ibl =A I
Av=AI=
Rcn (Ro)
R
i1
Rcn = Collector Ressistance of nth stage transistor
Ril = Input resistance of I stage amplifier
_
2 Rcn
Ap- (AI) . R.
I}
2.2.4
Choice of Transistor in a Cascaded Amplifier Configuration
By connecting transistor in cascade, voltage gain gets multiplied. But what type of configuration
should be used? Common Collector (CC) or Common Base (CB) or Common Emitter (CE)? To get
voltage amplification and current amplification, only Common Emitter (CE) configuration is used.
Because for Common Collector, the voltage gain is less than 1 for each stage. So the overall amplification
is less than 1.
Common Base Configuration is also not used since Al is less than 1.
RL
A =A x v I R.1
Effective load resistance RL is parallel combination of Rc and Ri of the following stage, (next
stage) (since in multi stage connection, the output of one stage is the input to the other stage). This
parallel combination is less than R i. Therefore, R L < 1.
Rl
The current gain AI in common base configuration is hib < 1 or == 1.
Therefore overall voltage gain == 1. Therefore Common Base configuration is not used for
cascading.
So only Common Emitter configuration is used. (hfe >> I)
Therefore overall voltage gain and current gains are> 1 in Common Emitter configuration.
For a single stage Resistor Capacitor coupled amplifier, bandwidth is (fi - fi) == fi. fi is very small
compared to fi. fi is usually of the order of KHz or MHz. fl is of the order of few HZ.
Electronic Circuit Analysis
46
n
n
Ifhn is the upper 3 db frequency
h is the frequency at which
_ AM
AH -
.J2'
When n stages of RC coupled amplifiers are connected in cascade, the upper 3db frequency
(h)n for which the overall voltage gain falls to
n
1
J2
of its midband value is,
Multistage Amplifiers
or
If
then
47
!If;;IT~
{If;; rr ~2
h
48
Electronic Circuit Analysis
I
I+
[1LJ2
fin
When n stages are connected in cascade, the lower cut off frequency for nth stage is,
n
1+ [
iLJ
2
fIn
or
[I+UJT' =h
or
1+(A]'=2V'
fin
fi =~21/n_l
fin
or
fiIn
-
-~~2l/n -I
-
49
Multistage Amplifiers
2.2.5
Midband frequency fo
It is the geometric mean of the lower cut off frequency 1\ and upper cut offfrequency ii. It lies in the
to = ~It f2 .
middle of the mid frequency range and has maximum gain . .
I~
and IT are the frequency constants of a Transistor in Common Emitter configuration.
For a Transistor in Common Emitter configuration, the plot of AI Vs RL is as shown Fig. 2.6.
Short circuit current gain of Common Emitter Transistor varies with frequency. Short circuit current
gain of transistor varies with frequency.
AI
i
----. R
L
Fig. 2.6 Variation of Ai with RL
The frequency fp; It is the frequency at which short circuit current gain of the Transistor
falIs to
121
of its maximum value. This is called as the Bandwidth of the transistor.
I
hje . I~
2.2.6
=
IT
I
Cascading Transistor Amplifiers
When the amplification of a single transistor is not sufficient for a particular purpose (say to deliver
output to the speaker or to drive a transducer etc) or when the input or output impedance is not of the
correct magnitude for the desired application, two or more stages may be connected in cascade.
Cascade means in series i.e. The output of first stage is connected to the input of the next stage.
II Stage
Fig. 2.7 Cascaded amplifier stages
Let us consider two stage cascaded amplifier. Let the first stage is in common emitter
configuration. Current gain is high and let the II stage is in common collector configuration to provide
high input impedance and low output impedance. So what are the expressions for the total current
50
Electronic Circuit Analysis
gain AI of the entire circuit (Le. the two stages), Zi, Av and Yo? To get these expressions, we must
take the h-parameters of these transistors in that particular configuration. Generally manufactures
specifY the h-parameters for a given transistor in common emitter configuration. It is widely used
circuit and also AI is high. To get the transistor h-parameters in other configurations, conversion
formulae are used.
2.2.7
The Two Stage Cascaded Amplifier Circuit
The Transistor Q) is in Common Emitter configuration (Fig. 2.8). The second Transistor Q2 is in
Common Collector (CC) configuration. Output is taken across 5K, the emitter resistance. Collector is
at ground potential in the A.C. equivalent circuit.
Biasing resistors are not shown since their purpose in only to provide the proper operating point
and they do not affect the response of the amplifier. In the low frequency equivalent circuit, since the
capacitors have large value, and so is Xc low, and can be neglected. So the capacitive reactance is not
considered, and capacitive reactance Xc
1
=
2rcf C is low when C is large and taken as short circuit.
The small signal Common Emitter configuration circuit reduces as shown in Fig 2.8 .
. . . - : : - - - - - - - - - , - - + vee
Fig. 2.8 Two stage cascaded amplifier circuit
In this circuit Q2 collector is at ground potential, in AC equivalent circuit. It is in Common
Collector configuration and the output is taken between emitter point E2 and ground. So the circuit
is redrawn as shown in Fig. 2.9 (a) indicating voltages at different stages and input and output
resistances.
Re2 5Kn
N
Fig. 2.9 (a) Redrawn circuit
51
Multistage Amplifiers
2.2.8
Phase Response
Since the transistor is in Common Emitter configuration, there wiJI be a phase shift of
input and output.
But in the L.F. range,
e = + tan- I
Isbo between
[))
Whenf= fi, e = + tan -I (1) = 45°. As/, increases, e decreases.
Total phase difference in the L.F. range is, + IS00. Phase shift due to Common Emitter
configuration
e = + tan- l (~)
This phase lead decreases as frequency increases.
In the high frequency range
e = tan-i~).
When f= f 2,
e = tan- i
1 = 45°
As f increases, Bdecreases. Therefore Phase lag decreases
In the low frequency range, the signal is leading. Since
Total Phase angle = + IS00 +
e;
<I>
e is positive and + IS00,
= IS00 + tan- i
In the high frequency range the signal is lagging. Therefore + IS00 <I>
= IS00 - tan- i
(9)
e;
~)
Phase Response: IS00, because, Common Emitter configuration introduces tSOO phase shift.
<I>
IS00 + e
i
180 + 45°
180°
Fig. 2.9 (b) Variation of phase angle with frequency
In the low frequency range, Phase shift ¢ = (1800 + B)
In the high frequency range, Phase shift ¢ = (J 800 - B)
2.2.9
Gain - Bandwidth Product
AMF=
-hfe[RcRL]
RC+RL
~
Bandwidth = J;
- fi :::
fi
fi «J;
Electronic Circuit Analysis
52
21tC S ·
' RC-RL
[ RC+RL
1
The product of these two, (AMF and BW) is,
xii = fT =
-h fe ... 1
-h-.- x 21tC
=fT
S
For a given value of CS' this product is constant.
We can increase the voltage gain by increasing R c ' RL parallel combination.
A
le
But
fi
ex:
1
. Therefore
RC IIRL
fi
will reduce if AMF is increased. Therefore
if voltage gain
increases, Bandwidth decreases and vice versa.
fi can be increased without affecting voltage gain by reducing the value of Cs - But there is a
limit to which Cs can be reduced. If higher gain is required then Bandwidth has to be scarificed.
It (B.W) will reduce.
The product ofmidband gain and Bandwidth is also known as the figure ofMerit ofthe circuit.
Example : 2.5
If ~ = I SO, what are the cutoff frequencies of the input and output lead networks of the
given circuit?
Fig. 2.10 Circuit for Ex : 2.5
Solution
When ~ value is given, and not hie of the transistor, the input impedance of the transistor can
be determined.
Yin = iere'
V in :::: ~ i b· re'
53
Multistage Amplifiers
Rin = Rl II R211 Pre'
Pre' = 150 x 22.7 = 3.14 KQ
Rin = 10K 112.2K 113.14 K = 1.18KQ
.
1
Ii = - - - - m
21t(Rs + Rin )C in
Cutoff frequencies of input network (HPF)
1
21t(Ro + RL)C O
10=----1
lin =
=
21t(1KQ+1.18KQ)(0.47IlF)
155Hzs.
10
21t(3.6K + 1.5 KQ) (2.2 IlF)
14.2 Hzs
LPF Cutoff frequency of output network.
=
2.2.10 Emitter Bypass Capacitor
Fig. 2.11 CE Amplifier circuit
CE is the emitter bypass capacitor. This causes the frequency response of an amplifier to break
at a cutoff f!'equency, designated IE' To understand the effects of emitter bypass capacitor,
suppose, Cin and Co (coupling capacitor) are shorted, then the frequency response will be, as shown
in Fig. 2.12(a).
Am
0.707 Am
~
(a)
(b)
Fig. 2.12 Effect of emitter by pass capacitor
Electronic Circuit Analysis
54
This means thrlt frequency response breaks at f E. Thevenin's equivalent resistance driving
Common Emitter, Rout is the Thevenin's resistance facing the capacitor.
'+ RsIIRIIIR2
Rout ::: re
/3
[ie r~ + ieRE - V in + ib (RsIIRIIIR2)} = 0
':ib=ic / /3, :ie / /3,
.
V
,
In
RE +re +(Rs IIR I IIR 2)//3
The emitter resistor RE is driven by an AC source with an AC output resistance of
Solving for ie'
Ie :::
Z (emitter)
o
2.3
= r' +
Rs II Rill R2
e
/3
Equivalent Circuits
I
r
+
_R",--S_IIR-cI,-II_R..::...2
e
J3
Fig. 2.13 Equivalent circuits
1
fE= - - - 21tR
C
out E
fE
= Cutoff frequency of emitter network.
Rout = Output resistance facing bypass capacitor.
= Emitter bypass capacitance.
CE
!o3.1
Decibel
n many cases it is convenient to compare two powers on a logarithemic scale rather than on a linear
;cale. The unit of this logarithemic scale is called the decibel abbreviated as db.
Suppose P2 is the output power and P I is the input power, then the power gain in decibels is
N
=
10 loglo
P2
pI
Where N is in dbs
If N is negative, it means P2 is less than PI'
Noise power is also expressed in decibels (dbs). It should be negative for a given device or
amplifier i.e the output noise is less than what is present in the input.
Multistage Amplifiers
55
If for a given amplifier circuit, the input and output resistances are same (as R), then
V
2
~ where V 1 and V2 are input and output voltages.
P2 =
R
N
=
20 log
V
---.l.
=
20 log Av
VI
But even though, the input ana output impedances are not equal, this cqI\vention is followed for convenience i.e, N = 20 log Av' If Av = 10, N = 20 log lOis 20 the d~ibel voltage gain of the
amplifier. 20 is not the power gain because, the input resistances are n~l equal. Therefore 20 is the
decibel voltage gain. If the output resistances are equal decibel voltage gain is equal, to power gain.
Overall db V of a multistage amplifier is equal to sum of db V of individual stages.
2.4
Miller's Theorem
Fig. 2.14(a) shows an amplifier with a capacitor between input and output terminals. It is called as
feedback capacitor. When the gain K is large, the feedback will change the input Z and output Z of
the circuit.
e
,.-----11 (1------.
:~vm~'I
~;v~
K-----I11---1
L---
Fig. 2.14 (a) Feedback capacitor
A circuit as shown above is difficult to analyze, because of capacitor. So according to the
Miller's theorem, the feedback capacitor can be split into two values, one as connected in the input
side and the other on the output side, as shown in Fig. 2.14 (b).
Fig. 2.14 (b) Splitting of feedback capaCitor using Miller's Theorem.
2.4.1
Mathematical Proof of Miller's Theorem
-.
The AC current passing through capacitor (C) in Fig. 2.14 (a) is
Ie
=
Yin - Vout - (VIn - Vout)
(_.
= - J Xc
1_)
Jroc
Vout = K Yin
Electronic Circuit Analysis
56
Vin (l-K)
-jXe
Vin (1- K)
-jXe
=
-jXe
(l-K)
-j
21tfC(1-K)
Since
1
Xc
=
2rcfC
Vm is the input Z as seen from the input terminals.
Ie
Zin
=
-j
21tf[C(l-K)]
C in = C (l-K)
Similarly output capacitance can be derived as follows:
Current in the capacitor,
V(l-~)
out
V
- j Xc
- jXe
Vout
-1-
e
= Zout
Vout(l-k)
-jXc
-jXe
-j
(A;l)
C out (Miller) = C
(KK-l)
out
57
Multistage Amplifiers
2.5
Frequency Effects
2.5.1
Lead Network
Fig. 2.15 Lead Network
Vm
I=--..!!!.I
R+-jroC
Vin·R
Vo
I. R =
=
(
I)
R+jroC
Vo leads with respect to Yin' So it is called as lead network for the above circuit, at low
frequencies, Xc = 00.
:. Vo is low since (I is low). As f increases Xc decreases. Hence I flows, and V0 increases.
:. Gain increases. Hence the frequency response is as shown.
2.5.2
Cut-off Frequency
Vin
0.707
i
Ie
--+1
Fig. 2.16 Frequency response
V
out
=
R
V
I 2
2 • in
R
V +Xc
Vout =
Yin
R
~R2+x~
Cut-off frequency is the frequency at which
V
YOU!
m
This happens when, Xc = R
I
--=R
rc
2 fc C
Lower cut-off frequency.
=
I
r;::.
,,2
58
2.5.3
Electronic Circuit Analysis
Half Power Point
2
At
V out
2R
P=--
f= fe-
Power is half of maximum power when Vout is maximum. Therefore it is also called as half
power point.
Fig. 2.17 Equivalent circuit
Considering Source Resistance:
V out
R
=
)(Rs+Rd+ X /
Vin
1
J;=---e
21t(Rs + Rd.C
In the midband frequency region Xe == O.
Vout =
RL
Vm
Rs +RL
Amid
2.5.4
(midband voltage gain) =
RL
RS+RL
---"'-
Stiff Coupling
If Xc = 0.1 (RS + R L) it is called Stiff Coupling.
The coupling capacitor ml.lst have (or bypass capacitor).
This is known as Stiff Coupling.
Multistage Amplifiers
2.6
59
Amplifier Analysis
,.-----,-----__<t + Vee
Fig. 2.18 Amplifier circuit
The equivalent circuit from input side, not considering Rs'
The equivalent circuit from output side, not considering R L .
Rin = R) II R2 I Pr~ (Not cosidering Rs)
Rout == Rc (Not cosidering RL )
Fig. 2.19 Equivalent circuit
Considering the input side as a load network,
1
1;n = 2n(Rs +Rm)C m
Similarly output lead network has cut-off frequency
I
2.6.1
Lag Networks
1
-,-,Vm
JroC
V =~===
o ~R 2 +Xc 2
V ==
o
- j(V. / roC)
~R2+X~
therefore it is lag network
Electronic Circuit Analysis
60
Vo lags with respect to Vi'
VOU!
Xc
=
Vin
~R2+Xc2
R
Fig. 2.20 Lag Network
Ie>
at
R=Xc
kc=~1
RL
Amid = R +R
s
L
Midband gain
2.6.2
Decibel
Power gain
P
=G = -2
PI
P2 = Output power.
PI = Input power.
Decibel power gain = G' = 10 loglo G.
If G = 100, G' = 10 log 100 = 20 db.
If G = 2, G' = 10 log 2 = 3.01db.
Usually, it is rounded off to 3.
2.6.3
Negative Decibel
If a < 1, a 'will be negative.
1
"2'
If
G=
If
I
G'= 10 log - =-3.01 db
2
G
G'
1
o db
10
10 db
100
20 db
30 db
1000
10,000 40 dh
Multistage Amplifiers
61
Ordinary Gains Multiply:
If two stages are there,
P2~
Fig. 2.22 Cascaded amplifier stages
G=.!2· G =.!l
1
p' 2
P
I
Overall gain
2
P3
G=PI
P
P
- 3
PI . P2
2
=-
G
=G1 Gz
Decibel gains add up.
2.6.4
Decibel yoltage Gain
If A = Normal Voltage Gain Y2, decibel voltage gain A' = 20 log A. A = yy2
Y
I
If Rl is output resistance,
y2
Input power
P
= _I
1
RI
If R2 is output resistance, P2 =
y2
~
2
:. Power gain
G= P2 =
PI
y/ . ~
y/ R2
If the impedances are matched, i.e. input resistance
Rl = output resistance ~...
Power gain
G
y2
y2
= _2_ =
A2
I
In decibels, 10 log G
=
G' = 10 log A2.
In decibels, 10 log G
=
G'
=
20 log A.
62
2.6.5
Electronic Circuit Analysis
Cascaded Stages
Fig. 2.23 Cascaded stages
A =AI x A2
Al = AI' + A 2 , (in decibels)
2.6.7
Stiff Coupling
When the capacitor is chosen such that, Xc =
-R
10E
,
it is called as stiff coupling. Bec,ause, for the
circuit shown.
c
0.707
~
I
Fig. 2.24 Stiff coupling network
For the circuit, shown above, in the mid frequency range, Xc is negligible.
.R
Vout =
L ;::
1.
(Vin)
Rs +RL
RL ·
A(mid) ;:: Voltage gain in the mid frequency range is,
V out
Amid ==
Vin ;::
. . Vont
For the complete cIrcUIt,
=
V.
tn
RL
Rs +RL
I
in
V
;::
2
V(Rs + Rd + X/
When
V out
RL
RL
.J2 (Rs + Rd
= 0.707 Amid
Multistage A'mpiifzers
63
!c
1
=
e
When
21t(Rs +RL).C
Xc
= 0.1 (Rs + R L),
RL
YOU!
=
Yin
~(RS+RL)2+[0.1 (Rs+Rd]2
YOU!
V.
= 0.995 Amid'
In
Xc is made
= 0.1 (Rs + R L), at the lowest frequency. ife lower)
.. At the frequency, A = 0.995 Amid' So it is called as Stiff Coupling.
R = -R- = 21tjRC = -I
Xc
1
Ie
21t Ie
-
fc =
1
21tRC
Decibel voltage gain A • ~ 20 log
Ie
When
I
~I+(:')'
= cutoff frequency.
Ie
= 0.1,
1
Al = 20 log
'/==
,,1 +(0.1)2
== Odb
Electronic Circuit Analysis
64
When
1
~ = - 3.01 db == 3 db
Al
=
20 log
AI
=
1
20 log - -
VI + 12
When
When
fe
L
'
=
20 log
=
20 log
fe
f=
1
~1+lO02
= -
40 db
1000
'
=
fe
Al
.. when
20 db
= -
L= 100
AI
When
~1+Hi
10'
f=fc '
f= 10 fe>
1
~1+10002
=
60 db
A' =0
Al
= -
3 db
Al
= -
20 db
and so on.
1000 Ie
I
-3db
- 20
- 40
- 60
Fig. 2.25 Frequency RolI- off
.. A change of 20db per decade change in frequency.
An Octave is a factor of 2 in frequency change
Whenf changes from 100 to 200 Hzs, it has changed by one octave.
Multistage Amplifiers
65
When,f changes from 100 to 400 Hzs, it is two octaves for lead network, Bode Plot is
0.001
-w~
___
- 40 db
- - -
I
I
~
Ie
0.01
____
I
I
I
~
Ie
I
___ J
I
-t - - - I
- 60 db
Fig. 2.26 Variation of Av with f
2.7
High Input Resistance Transistor Circuits
In some applications the amplifier circuit will have to have very high input impedance. Common
Collector Amplifier circuit has high input impedance and low output impedance. But its Av < 1.
If the input impedance of the amplifier circuit is to be only 500 KO or less the Common
Collector Configuration can be used. But if still higher input impedance is required a circuit shown in
Fig. 2.29 is used. This circuit is known as the Darlington Connection (named after Darlington) or
Darlington Pair Circuit.
r--------------,
..--~__.,.---"---__oc
B
B
E
c
PNP
I
"'--------------""
Fig. 2.27 Darlington Pair Circuit
In this circuit, the two transistors are in Common Collector Configuration. The output of the
first transistor Q I (taken from the emitter of the Q\) is the input to the second transistor Q2 at the base.
The input resistance of the second transistor constitutes the emitter load of the first transistor.
So, Darlington Circuit is nothing but two transistors in Common Collector Configuration connected in
series. The same circuit can be redrawn as AC equivalent circuit. So, DC is taken as ground shown
in Fig.2.29. Hence, 'C' at ground potential. ColIectors of transistors Q\ and Q 2 are at ground potential.
66
Electronic Circuit Analysis
The AC equivalent Circuit is shown in Fig. 2.29.
-Vee
c
II
B
VI
Fig. 2.28 AC equivalent circuit of Darlington Pair
There is no resistor connected between the emitter of Q I and ground i.e., Collector Point.
So, we can assume that infinite resistance is connected between emitter and collector. For the
analysis of the circuit, consider the equivalent circuit shown in Fig. 2.29 and we use Common
Emitter h-parameters, hie' hre • hoe and hfe .
B
1
Q,~
VI
1
R=OO
I
Fig. 2.29 Simplified circuit
For PNP transistor, Ic leaves the transistor, Ie enters the transistor and Ib leaves the transistor.
2.7.1
Current AmpliflCation for Darlington Pair
But
Ic = ICJ + IC2 '
Iq = Ibl h fe ;
Ic = Ib hfe · (Assuming identical transistor and hfe is same)
2
2
Ib = Ie (The emitter of Q I is connected to the base of Q2)
2
I
Ic = Ibl h te +Iel hfe
Ie = Ib + Ic = Ib (1 + hfe )
I I = hI I I
I
c
fe b l
l
..... (1)
..... (2)
67
Multistage Amplifiers
Substituting equation (2) in (I),
Ie
Ib h re + Ib (I + hre)h re = Ib (2hre + hre2)
I
I
I
=
Since, hre is of the order of 100.
:.
Ie = Ib h r/
I
It means that we get very large current amplification (AI =
~l' in the case of Darlington Pair
Ibl
Circuit, it is of the order hre2 i.e. l002 = lO,OOO.
2.7.2
Input Resistance (R)
Input resistance Ri2 of the transistor Q 2 (which is in Common Collector Configuration) in terms of
h- parameters in Common Emitter Configuration is,
Ri2
=
hie + (1 +hre ) RL
But
Here RL is Re, since, output is taken across emitter resistance.
Ri2 == (1 + hfe) Re
The input resistance Ril of the transistor Q I is, since it is in Common Collector Configuration
Ri = hie + hre AI·RL'
Expressing this in term of Common Emitte- h-parameters,
hic == hie; hre == l.
(For Common Collector Reverse Voltage Gain is equal to 1) and RL for transistor Q1 is the
input resistance of transistor Q2'
..
Therefore,
Ril
hie + All R i2 · R i2 , is large,
hoe .Ri2 .::: 0.1. and AI == 1+ hre
=
Ril == All Ri2
Ri2 = (1 + hfe) Re
But the expression for Common Collector Configuration in terms of Common Emitter
h-parameters is
A=
I
Here,
RL
=
1+ h fe
l+hoe.RL
Ri2
and
Ri2
=
(1 + hfe) Re'
68
Electronic Circuit Analysis
All
=
1+ h
fe - ----=
1+ hoeCl + hfe)Re
hoe Re will be is less than 0.1 and can be neglected .
.. ho value is of the order of Il mhos (micro mhos)
e
All
=
1+ h fe
1+ h oe h fe R e
Ri\ ::: All· Ri2
This is a very high value. If we take typical values, of Re = 4KW, using h-parameters,
Ri2 = 205 KO.
Ri = 1.73 MO.
AI = 427.
Therefore, Darlington Circuit has very high input impedance and very large current gain
compared to Common Collector Configuration Circuit.
2.7.3
Voltage Gain
General expression for Ay for Common Collector in term of h-parameters is
~;
Ri
Ay = 1 -
But
hie ==h ic or Ay\
=
Yy2 =[1- Rhi.e 1
I
II
Ri == All· R 2 ·
Ay\
=
(
-
hIe
All
Ri2
1
R12 »h·Ie and A.IJ is »
Therefore, Ay is always less than 1.
1
69
Multistage Amplifiers
2.7.4
Output Resistance
The general expression for Ro ofa transistor in Common Collector Configuration in terms of Common
Emitter h-parameters is,
ROJ ==
Rs + hie
1+ h fe
Now for the transistor Q2' Rs is ROJ '
Rs + hie
+h·
1 + hfe
Ie
--=--=-
Therefore, R02 is the output resistance of the Darlington Circuit.
Rs + hie
hIe
R02 == (1 + hfeY + 1 + h fe
This is a small value, since, 1+ h fe is » 1.
Therefore, the characteristic of Darlington Circuit are
1. Very High Input Resistance ( of the order of M.o).
2. Very Large Current Gain (of the order of 10,000).
3. Very Low Output Resistance (of the order offew .0).
4. Voltage Gain, Av < 1.
Darlington Pairs are available in a single package with just three leads, like one transistor in
integrated form.
2.7.5
Disadvantages
We have assumed that the h-parameters of both the transistors are identical. But in practice it is
difficult to make h-parameters depend upon the operating point of QJ and Q2' Since the emitter
current of transistor QJ is the base current for transistor Q2' the value of IC2 » ICJ
1. The quiescent or operating conditions of both the transistors will be different. hje value will
be small for the transistor QI. .: hje = (f/fJ.lb2 is less
CDfL make CIL997 is a transistor of Darlington Pair Configuration with hje == 1000.
2. The second drawback is leakage current of the first transistor Q I which is amplified by the
second transistor Q2
(-:
leI == fbi
70
Electronic Circuit Analysis
Hence overall leakage current is more. Leakage Current is the current that flows in the circuit
with no external bias voltages applied
(a) The h-parameters for both the transistors will not be the same.
(b) Leakage Current is more.
....
Darlington transistor pairs are in single package available with hje as high as 30,000
2.7.6
. Boot Strapped Darlington Circuit
The maximum input resistance of a practical Darlington Circuit is only 2 MQ. Higher input resistance
cannot be achieved because of the biasing resistors RI , R2 etc. They come in parallel with R j of the
transistors and thus reduce the value of R j • The maximum value of R j is only
h1
.
smce,
h·
ob IS
h
t e
ob
resistance between base and collector. The input resistance can be increased greatly by boot strapping,
the Darlington Circuit through the addition of Co between the first collector C I and emitter B 2.
What is Boot Strapping?
I'
v '"
R
R,=R
R,'=Rll-k
Boot strapping
Fig 2.31
Fig. 2.30
In Fig. 2.31, V is an AC signal generator, supplying current I to R. Therefore, the input resistance of
the circuit as seen by the generator is R.
V
= T = R itself. Now suppose, the bottom end of R is not at
ground potential but at higher potential i.e. another voltage source of KV (K< I) is connected between
the bottom end of R and ground. Now the input resistance of the circuit is (Fig.2.3I).
r
or
R·'
•
VR
=
R
=-V(l-K) I-K
=
(V-KV)
R
Multistage Amplifiers
71
I' can be increased by increasing V. When V increases KV also increases. K is constant.
Therefore the potential at the two ends of R will increase by the same amount, K is less than 1,
therefore R j > R. Now if K = 1, there is no current flowing through R (So V = KV there is no
VI
Fig. 2.32 Boot Strap Circuit
potential difference). So the input resistance R\ = 00. Both the top and bottom of the resistor
terminals are at the same potential. This is called as the Boots Strapping method which increases the
input resistance of a circuit. If the potential at one end of the resistance changes, the other end of R
also moves through the same potential difference. It is as if R is pulling itself up by its boot straps.
For CC amplifiers Ay < 1
0.095. So R j can be made very large by this technique. K = Av
I.
Ifwe pull the boot with both the edges of the strap (wire) the boot lifts up. Here also, if the potential
at one end ofR is changed, the voltage at the other end also changes or the potential level ofR3 rises,
=
=
as if it is being pulled up from both the ends.
For Common Collector Amplifier,
h.
R = _I_e_. A =1.
\ I-Av.'
v
Therefore, R j can be made large, since it is of the same form as
In the circuit shown in Fig. 2.32, capacitor Co is connected between C 1 and E2. If the input
signal changes by Vj' then E2 changes by Ay.V j (assuming the resistance of Co is negligible).
Therefore, _1_ is now effectively increased to
h~
h~O-A0
=400Mn
72
Electronic Circuit Analysis
2.7.7
AC Equivalent Circuit
The input resistance
v.
R
= _I
I
Ibl
=h fe h fe Re
1
2
Ifwe take h fe as 50, Re = 4Kn, we get R j as 10Mn. Ifa transistor with h fe = 100 is taken, R j
will be much larger. The value ofXco is chosen such that at the lower frequencies, under consideration
XCo is a virtual short circuit. If the collector C 1 changes by certain potential, E2 also changes by the
same amount. So C 1 and E2 are boot strapped. There is
1
h
ob
between BI and C 1.
Reff = h ob (1 - A v )
h re V cel
V,
N
Fig. 2.33 AC equivalent circuit
Direct short circuit is not done between C 1 and E2 . Since, DC condition will change, Xc is a
o
short only for AC signals and not for DC.
2.S
The CASCO DE Transistor Configuration
Vo =V2
1
Fig. 2.34 CASCO DE Amplifier (C.E, C.B configuration)
73
Multistage Amplifiers
The circuit is shown in Fig. 2.34. This transistor configuration consists of a Common Emitter Stage
in cascade with a Common Base Stage. The collector current of transistor Q) equals the emitter
current of Q2.
The transistor Q) is in Common Emitter Configuration and transistor Q 2 is in Common
Base Configuration. Let us consider the input impedance (h))) etc., output admittance (h22) i.e.
the h - parameters of the entire circuit in terms of the Ir - parameters of the
two transistors.
2.S.1
Input
IZI (h ll )
If V2 is made equal to 0, the net impedance for the transistor Q) is only h ib2 . But h ib, for a
transistor in common emitter configuration is very small == 20. We can conclude that the collector of
Q) is effectively short circuited.
..
hll == hIe
When V2 = 0, C2 is shorted. Therefore, hl2 = h ib2 · But h ib2 is very small. Therefore C) is
virtually shorted to the ground.
Ihi
2.S.2
=
hie
I
Short Circuit Current Gain (h 21 )
I'
I1
=
hfe
since h
fb
I
IE
= ~
74
2.8.3
Electronic Circuit Analysis
Output Conductance (h22)
Output Conductance with input open circuited, for the entire circuit is,
1
when II = 0, the output resistance of the transistor Q I is
h
oe
=40 KO. (Since Q I is in Common
Emitter configuration and hoe is defined with II = 0).
is the source resistance for Q2' Q 2 is in Common Base Configuration. What is the value ofRo of the
transistor Q 2 with Rs
40k
=
It is == IIhob itself.
Since, hoel is very large, we can say that I ~ = 0 or between E2 and ground there is infinite
impedance. Therefore, output conductance of the entire circuit is h22
2.8.4
=hob'
Reverse Voltage Gain
,
h re ·
~
2
= h rb
(Since, Q 2 is in Common Base configuration)
hl2 == ~e hrb ·
hre
= 10-4
hrb = 10--4. .; hl2 is very small
hi = hll == hie'
Typical value = 1.1 KO
h f = h21 == h fe ·
Typical value = 50
ho = h22 == hob'
Typical value = 0.49 IlAIV
hr = hl2
Typical value = 7 x 10 - 8.
=h re hrb ·
75
Multistage Amplifiers
Therefore, for a CASCODE Transistor Configuration, its input Z is equal to that of a
single Common Emitter Transistor (hie)' Its Current Gain is equal to that of a single Common
Base Transistor (h fe ). Its output resistance is equal to that of a single Common Base
Transistor (hob)' The reverse voltage gain is very very small, i.e., there is no link between
V I (input voltage) and V 2 (output voltage). In otherwords, there is negligible internal feedback
in the case of, a CASCODE Transistor Circuit, acts like a single stage C.E. Transistor
(Since hIe and h fe are same) with negligible internal feedback (:. h re is very small) and very
small output conductance, (== hob) or large output resistance (== 2MO equal to that of a
Common Base Stage). The above values are correct, if we make the assumption that hob RL
< 0.1 or RL is < 200K. When the value of RL is < 200 K. This will not affect the values of
hI' hr' ho' h f of the CASCODE Transistor, since, the value of hr is very very small.
CASCODE Amplifier will have
I. Very Large Voltage Gain.
2. Large Current Gain (hje ).
3. Very High Output Resistance.
Example : 2.6
Find the voltage gains Ays' A YJ and AY2 of the amplifier shown in Fig.2.35.
Assume
and
hie
=
1KO,
hre
hoe
=
10 -8 A/V.
=
10-4,
Fig. 2.35 Amplifier circuit Ex : 2.6
76
Electronic Circuit Analysis
Solution:
The second transistor Q2 is in Common Collector Configuration QJ is in Common Emitter Configuration.
It is convenient if we start with the II stage.
II Stage:
hoe. RL2 = 10 - 8 x 5 x 103 = 5 x 10 - 5 < O.l.
Therefore, hoe RL2 is < 0.1, approximate analysis can be made, rigorous expression of C.C.
and C.B Configuration need not be used.
Ri2
+ (I +h fe )R L2
= hie
=
1Kn + (1 +50) 5Kn
=
256 Kn.
Av 2 is the expression for Voltage Gain of the transistor in Common Collector Configuration in
terms of Common Emitter h-parameters is
=
1-
=1- 0.0039 =0.996
lKn
256Kn
I Stage:
RLJ
= 10 K II
Ri2
= 10 K I 256 Kn
RiJ
",I
9.36 Kn .
.: approximate equation can be used
hoe· RLJ < O.l.
All = -
=
50
= hee · = lKn.
= _
R LI
hfe h·Ie
Overall Voltage Gain = A YJ . AY2
AyS =A y x
= -
= - 50 x 9.63 k = -48
482
lKn
x
0.996
= -
.
480.
Rs
IKn
= AV x--= -240.
Rs+h je
2K
Multistage Amplifiers
2.9
77
CE - CC Amplifiers
This is another type of two-stage BJTamplifier. The first stage in Common Emitter (CE) configuration
provides voltage and current gains. The second stage in Common-Collector (CC) configuration
provides impedance matching. This circuit is used in audio frequency amplifiers. The circuit is
shown in Fig. 2.36.
Fig. 2.36 CE, CC Amplifiers
Analysis
Here biasing resistors are neglected for simplification of analysis.
II Stage in CC amplifier :
RL2 ::: RL
hoc RLI.:s 0.1
AI2
=
Ri2
=
(1 + hfe )
(1 + hfe ) RL2
AI'2
_
=
_ A I2 ·R L2
--=--------''''R·
AY2 - A y2 ' -
'2
(l+h fe R L2 )
hie
+(1 +h fe )R L2
h·
=1-~'
R·
'2
'
AY2 < 1
for CE stage
Al = AI I = -hfe
I
I
Ril = Ril' = hie
Ay = Ay I = (AI . RL / Ri)
I
I
1
I
78
Electronic Circuit Analysis
Overall Characteristics
2.10
Two Stage RC Coupled JFET amplifier (in Common Source (CS) configuration)
The circuit for two stages of RC coupled amplifier in CS configuration is as shown in Fig. 2.3 7.
V~
Cb
t-----1
V
Cs
~I
Fig. 2.37 Two stage RC coupled JFET amplifier
The output Vo of I Stage is coupled to the input Vi of II Stage through a blocking capacitor
Cb. It blocks the DC components present in the output orI Stage from reaching the input of the II
stage which will alter the biasing already fixed for the active device. Resistor Rg is connected between
gate and ground resistor Ro is connected between drain and VDO supply. C S IS the bypass capacitor
used to prevent loss of gain due to negative feedback. The active device is assumed to operate in the
linear region. So the small signal model of the device is valid.
Frequency Roll-off is the term used for the decrease in gain with frequency in the upper
cut-off region. It is expressed as db/octave on db/decade. In the logarithmic scale of frequency,
octave
. 12
I~
IS
= 2 decad'
e IS
I~ = 10 .
12
The purpose of multistage amplifiers is to get large .gain. So with BJTs, Common Emitter
Configuration is used. If JFETs are employed, common source configuration is used.
2.11
Difference Amplifier
This is also known as differential amplifier. The function of this is to amplify the difference between
the signals. The advantage with this amplifier is, we can eliminate the noise in the input signals which
is common to both the inputs. Thus SIN ratio can be improved. The difference amplifier can be
represented as a blackbox with two inputs V 1 and V2 and output V0 where V0 = Ad (V 1 - V2)'
Multistage Amplifiers
79
where Ad is the gain of the differential amplifier. But the above equation will not correctly describe the
characteristic of a differential amplifier. The output Vo depends not only on the difference of the two
signals (V I - V2) = Vd but also on the average level called common mode signal Vc = V I + V2 I 2.
If one signal is (VI) 100 IlV and the other signal (V2) is -100 IlV.
..
Vo should be Ad (200) IlV.
Now in the second case, if V I = 800 Il V, and V2 = 600 IJ V. Vd = 800 - 600 = 200 IJV and Vo
should be Ad (200) IlV. So in both cases. for the same circuit. VI) should be the same. But in practice
it will not be so because the average of These two signals V I & V2 is not the same in both the cases.
Vd =V I -V2
1
V = - (V + V)
c
2
I
2
from the equations above, we can write that,
1
V I =V c + -2 Vd
[.,' If we substitute the values of Vc and Vd we
get the same.
I
VI
VI
V =V +- V = - + - = V ]
I e 2d
22
I
V0 can be represented in the most general case as
Vo = A V +A2 V2
Substituting the values of V I and V2
1
V() = A 1 [Vc +
1
"2 vd] + A2 [Vc - "2
Vd]
A1
A2
=AI VC+- Vd +A 2 Vc - - Vd
2
2
Vo = Vc (AI + A2) + Vd [
AI -A2 l
2 --'
V0 = VcAe + Vd Ad
where
for operational amplifiers, always input is given to the inverting node to get
that Ad is very large and Ac is very small.
Al and A2 are the voltage gains of the two amplifier circuits separately.
The voltage gain from the difference signal is Ad'
A t -(-A 2 )
so
2
Electronic Circuit Analysis
80
The voltage gain from the common mode signal is Ac'
V0 = Ad Vd + Ac Vc'
To measure Ad' directly set VI = - V2 = 0.5V so that
Vd =0.5-(-D.5)= IV.
V =
c
(0.5-0.5)
2
=0
V0 = Ad' I == Ad it self
If VI = - V2 and output voltage is measured,
Output voltage directly gives the value of Ad'
Similarly if we set VI = V2 = I V. then
V=
VI + V2
c
2
2
==-=IV.
2
Vo = 0 + Ac . I = Ac'
The measured output voltage directly gives Ac. We want Ad to be large and Ac to be very
small because only the difference of the two signals should be amplified and the average of the signals
Ad
should not be amplified. :. The ratio of the these two gains p =
Ac
is called the common mode
rejection ratio. This should be large for a good difference amplifier.
VO=Ad Vd+Ac Vc
Ad
A=c
p
2.12
Circuit for Differential Amplir.2r
In the previous D.C amplifier viz., C.B, C.C and C.E, the output is measured with respect to ground.
But in difference amplifier, the output is a. to the difference of the inputs. So Vo is not measured w.r.t
ground but w.r.t to the output of one transistor Q 1 or output of the other transistor Q2'
Multistage Amplifiers
81
+
vee
BI
B2
EI
E2
R
81
Re
VI
Egl
V2
-VEE
I
E
I
82
Fig. 2.38 Differential amplifier
Equivalent Circuit
h.e
h.e
2
Bl
Ie
2
Neglecting hre & Ilh oe' 1lhoe
2
Fig. 2.39
.
The advantage with this type of amplifiers is the drift problem is eliminated. Drift means, even
Equivalent circuit
when there is no input, Vi there can be some output Vo which is due to the internal thermal noise of
th~ circuit getting amplified and coming at the output. Drift is reduced in this type of circuit, because,
the two points should be exactly identical. Hence, Ie ' hFE' VBE will be the same for the two transistors.
Now if Ie rises across RL (Ie R L ) increases with hu;rease in Ie . So the voltage at collector of Q 1
decreases~ If Q2 is also identlcallo Q1 its collectQr voltage also Idrops by the same amount. Hence
Vo which is the difference of these voltages remains' ~he same thus the drift of these transistors gets
cancelled.
The input to a differential amplifer are of two types. I. Differential mode 2. Common mode.
If V 1 and V2 are the inputs, the differential mode input = V2 - V I'
Electronic Circuit Analysis
82
Hefe two different a.c. signal are being applied VI & V2' So these will be interference of these
slgnab and so both the signals will be present simultaneously at both input points i.e., if V 1 is applied
at point I, It also prices up the signal V 2 and so the net input is (V I + V 2)' This is due to interference.
Common node input =
VI +V2
--!.---'=-
2
An ideal differential amplifier must provide large gain to the differential mode inputs
gain to command input.
and
zero
..... (1 )
A2 = voltage gain of the transistor Q2
Al = voltage gain of the transistor Q I
we can also express the output in term of the common mode gain Ac and differentil gain Ad'
..... (2)
..... (3 )
A
VO=V2(Ad+ 2C)
Comparing
eqn~.
-Vl(Ad-~c-)
4 and 1,
Ac
A =A + 2
d
2
Ac
A=A + I
d
2
Solving these two eqns. Ad =
Al +A2
----'---=
2
Ac=A2 -A1/2
I..IA 730 is an I.C ditferential amplifier.
S
6
7
8
Fig. 2.40 Ie ,uA 730 Pin c01figuration
Multistage Amplifiers
83
8 pins. Input is given to pins 2 and 3. V+ cc to pin 7, 4 is ground. Output is taken at pin no 6.
In the difference amplifier, the difference of the input voltages V j and V2 is amplified. The collectors
of the transistor Q j & Q 2 are floating. They are not at ground potential. So the output voltage is not
at ground potential. Hence the output voltage is the difference of the collector voltages (a.c) of
transistors Q j & Q2. Difference amplifiers are used in measuring instants and instrumentations
system~. The difference of Vi & Vi may be 1 IlV which is difficult to measure. So if this is
amplified to 1mV or 1V the mleasure~ent will be accurate. So difference amplifiers are used to
measure very small increased voltages.
While computing AI & A2 of individual transistors, the other input should be made zero. One
while computing AI' V2 = O. Because there should be no common mode signal, while computing A"
AI is the actual gain. Not differential gain. :. the other input is made zero.
In the case of operational amplifiers for single ended operation, always the positive end is
grounded (non inverting input) & input is applied to the inverting input (-). It is because, at this part
the feedback current & input current get added algebraically. So this is known as the swimming
junction. When sufficient negative feedback is used, the closed looop performance becomes virtually
independent of the characters of the operational amplifiers & depends on the external passive elements,
which is desired.
Electronic Circuit Analysis
84
1.
Based on·the type of Coupling, the Amplifiers are classified as _ _ _ _ _ _ _ __
2. Based on Bandwidth, the amplifiers are classified as _ _ _ _ _ _ _ __
3. Different types of Distortion in amplifiers are _ _ _ _ _ _ _ __
4.
When n stages with gains A\, A2 ....... An are cascaded11 overall voltage gain
AVn= _____________
5.
Expression for voltage gain Ay in the Mid Frequency range, interms of hfe ,
Av(M.F) = - - - - - - -
6.
Expression for Ay (H.F) in terms of/,Iz and Ay (M.F) is, Ay (H.F)
7.
Phase shift
8.
Expression for Ay (L.F) in terms of/'I; and Ay (M.F) is __________
9.
When n stages are cascaded, the relation betweenJ;n' andJ; is ____________
q, interms ofI;
1\, Rs and hIe is,
= _ _ _ _ _ _ _ __
and f is, _ _ _ _ _ _ _ __
10. When h stages are cascaded, the relation betweenl;nandl; is, ___________
11. The relation between h fe , fi3 and
~
is
12. Figure of Merit of an amplifier circuit is
13. CMRR(p)
=
14. Expression for Vo interms of Ad' Vd, Vc and pis,
15. Expression for Ad and Ac interms of At and A2 are
16. Stiff coupling is
17. When frequency change is octave
18. In cascade form, ordinary gains
~~
=
decibal gains
19. Phase response is a plot between
20. According to Miller's theorem the feedback capacitance when referred to input side, with gain
Ais _ _ _ _ _ _ _ _ __
21. Interms ofh fe , current gain in Darlin~on Pair circuit is approximately _ _ _ _ _ _ _ __
22. The disadvantage of Darlington pair circuit is _ _ _ _ _ _ _ _ __
85
Multistage Amplifiers
23. Compared to Common Emitter Configuration R, of
Darlington pair circuit is
24. In CASCODE amplifier, the transistors are in
configuration.
25. The salient featuers of CASCODE Amplifier are _ _ _ _ _ _ _ __
26. What is distortion ?
27. What are the types of distortion ? Define them.
28. How does the amplifier behave for low frequencies and high frequencies?
29. Which configuration is the best in cascade for an output stage and for an intermediate stage?
30. What is the darlington pair? What is its significance?
3 1. How is the bandwidth of a cascade affected compared to the bandwidth of a single stage?
32. Why is the emitter bypass capacitor used in an RC coupled amplifier?
33. What is the affect of emitter bypass capacitor on low frequency response?
34. What are types of cascade?
35. How would you differentiate an interacting stage from a non-interacting stage?
36. What is the expression for the upper 3dB frequency for a n-stage non interacting cascade?
37. What is the expression for lower 3dB cutoff frequency in n-stage interacting cascade?
38. What is the expression for upper 3dB cutoff frequency in a n-stage interacting cascade?
39. What is the slope of the amplitude response for an n-stage amplifier?
40. Why do we go for multistage amplifier?
Electronic Circuit Analysis
86
1.
Explain about the classification of Amplifiers based on type of coupling and bandwidth.
2.
What are the different types of distortions possible in amplifiers outputs?
3.
Obtain the expression for the voltage gain Av in the L.F, M.F and H.F ranges, in the case of
single stage BJT amplifier.
4.
When h-identical stages of amplifiers are cascaded, derive the expressions for overall gain Avn
lower cu~off frequency /;n and upper cutoff frequency h.n·
'
5.
With the help of necessary waveforms, explain about the step response of amplifiers.
6.
What is the significance of square wave testing in amplifiers?
7.
Draw the circuit for differential amplifier and derive the expression for CMRR.
8.
Explain about the characteristics of operational amplifiers.
9. Draw the circuit for Darlington pair and device the expressions for AI' Av' R, and Ro'
10. Draw the circuit for CASCODE Amplifier. Explain its working, obtaining overall values of the
circuit for h" hI' ho and h,
Multistage Amplifiers
87
1.
(a) R-C coupled
(c) Direct coupled
(b) Transformer coupled
(d) Tuned circuit
2.
(a) Narrow band
(b) Untuned amplifiers
3.
(a) Frequency distortion
(b) Phase distortion
4.
Avn = Ai' A 2, A3 ..... An
5. Av (M.F) = -hfe .R L
RS +hie
6. A, (H.F)
7.
~ Av (MFYI +{fz)
~ ~ 180" -
tan-I (
1)
8. A, (L.F) ~ AVrFj
1- j ]
9.
12n =
12
10. fIn =
11. hfe,f~
~~l/n -1)
XJ2Yn __)
=
IT
12. Figure of merit = Gain x Bandwidth
13. CMRR=(Ad/A c)
14. VO=Ad Vd (1+!...VC]
P Vd
15. Ad=.
(AI +A 2 )
2
; Ac=(A2-A)/2
16. When capacitor is chosen such that Xc =
17. 2
18. Multiply, Add up
R
1;
(c) Amplitude distortion
Electronic Circuit Analysis
88
19. Phase angle and frequency
20. Cm=C(l-A)
21. (hfe~
22. Leakage current is more
23. Very high
24. CE-CB
25. Large voltage and current gains
26. If the 04tput waveform is not the replica of the input waveform, it is called distortion.
27.
(i) Non linear c1istortion : If harmonic frequencies are generated at the o/p, it is called non
linear or amplitude distortion.
(ii) Freqeuncy distortion: If different frequency components are amplified differently, it is
called frequency distortion.
(iii) Phase distortion: If the output is shifted by different phased each time, it is called phase
distortion.
28. Low frequencies - high pass filter.
High frequencies - low pass filter.
29. Output stage - CC
Intermediate stage - CE
30. CC-CC cascade is called Darlington pair.
Significance: (i) It has very high input impedance. (ii) It behaves life a constant current source.
3 1. It decreases.
32. To decrease the loss in gain due to negative feedback
33. The tilt is more
34. Interacting and non interacting.
35. If the input impedance of next stage loads the previous stage, it is calleajnteracting stages. If
the input impedance of next stage does not load the previous stage, it is called non interacting
stage.
36.
37.
38.
f
A =~.J 1n/
2 -1
IH = f .J21/n-1
IH = 0. 94 / D •
39. 6n dB/octave or 20n dB/decade.
40. For high gain.
UNIT - 3
High Frequency
Transistor Circuits
In this Unit,
• Single stage amplifiers in the three configurations of C.E, C.B, C.C,
with design aspects are given.
• Using the design formulae for ~, AI' R Ro etc, the design of single
stage amplifier circuits is to be studied.
I
,
• Single stage JFET amplifiers in C.D, C.S and C.G configurations are
also given.
• The Hybrid - 1t equivalent circuit of BJT, expressions for Transistor
conductances and capacitances are derived.
• Miller's theorem, definitions for~ andh are also giv~n.
• Numerical examples, with design emphasis are given.
90
3.1
Electronic Circuit Analysis
Transistors at High Frequencies
At low frequencies it is a~sumed that;'ransistor responds instantaneously to changes in the input
voltage or current i.e., if you g~~~l between the base and emitter of a Transistor amplifier in
Common Emitter configuraii6n and ifthe input signal frequency is low, the output at the collector
will exactly follow the change~Hf the input (amplitude etc.,). If '1' of the input is
high (MHz) and the amplitude of the input signal is changing the Transistor amplifier will not be able
to respond.
What is the reason for this ? It is because, the carriers from the emitter side will have to be
injected into the collector side. These take definite amount of time to travel from Emitter to Base,
however smalt-i'rmay be. But if the input signal is varying at a much higher speed than the actual time
taken for the carries to respond, then the Transistor amplifier will not respond instantaneously. Thus,
the junction capacitances of the transistor, puts a limit to the highest frequency signal which the
transistor can handle. Thus depending upon doping area of the junction etc, we have transistors
which can respond in AF range and also RF range.
To study and analyze the behavior of the transistor to high frequency signals an equivalent
model based upon transmission line equations will be accurate. But this model will be very
complicated to analyze. So some approximations are made and the equivalent circuit is simplified. If
the circuit is simplified to a great extent, it will be easy to analyze, but the results will not be accurate.
If no approximations are made, the results will be accurate, but it will be difficult to analyze. The
desirable features of an equivalent circuit for analysis are simplicity and accuracy. Such a circuit
which is fairly simple and reasonably accurate is the Hybrid-pi or Hybrid-;rmodel, so called because
the circuit is in the form of 7r. The parameter have units of fl, U etc. So it is a hybrid circuit. Hence
it is called as hybrid-;r model. Using this model a detailed analysis of single stage Common Emitter
Transconductance amplifier is made.
3.2
Hybrid - x Common Emitter Transconductance Model
For Transconductance amplifier circuits Common Emitter configuration is preferred. Why? Because
for Common Collector (hrc < 1). For Common Collector Configuration, voltage gain Av < 1. So even
by cascading you can't increase voltage gain. For Common Base, current gain hib < 1. So overall
voltage gain is < 1. But for Common Emitter, hre» 1. Therefore Voltage gain can be increased by
cascading Common Emitter stage. So Common Emitter configuration is widely used.
The Hybrid-x or Giacoletto Model for the Common Emitter amplifier circuit (single stage)
is as shown :
r
B
1
b'c
=gb'c
VVv---~---.~~~r------~._---r----~C
Eo---------~~~~~~._------~--~----~E
Fig. 3.1 Hybrid
-1t
C.E BJT Model
High Frequency Transistor Circuits
91
Analysis ofthis circuit gives satisfactory results at all frequencies not only at high frequencies
but also at low frequencies. All the parameters are assumed to be independent of frequency.
+-- Ie
B
PNP
+
E --------~~----------~--------~
Fig. 3.2 PNP transistor amplifier
3.2.1
Circuit Components
B' is the internal node of the base of the Transconductance amplifier. It is not physically accessible.
The base spreading resistance rb'b is represented as a lumped parameter between base B and internal
node B'. (~ Vb'e) is a current generator. Vb'e is the input voltage across the emitter junction. If
Vb'e increases, more carriers are injected into the base of the transistor. So the increase in the
number of carriers is
(l
Vb'e. This results in small signal current (since we are taking into account
changes in Vb'e). This effect is represented by the current generator ~ Vb'e. This represents the
current that results because of changes in Vb'e' when C is shorted to E.
When the number of carriers injected into the base increase, base recombination also increases.
So this effect is taken care of by gb'e. As recombination increases, base current increases.
Minority carrier storage in the base is represented by Ce' the diffusion capacitance.
According to Early Effect, the change in voltage between Collector and Em!tter changes
the base width. So base width will be modulated according to the voltage between Collector and
Emitter. When base width changes, the minority carrier concentration in base changes.
Hence the current which is proportional to carrier concentration also changes. So IE
changes and hence Ie changes. This feedback effect [IE on input side, Ie on output side] is taken
into account by connecting gb'e between B', and C. The conductance between Collector and
Base is gee.
Ce represents the collector junction barrier capacitance.
3.2.2
Hybrid - n Parameter Values
Typical values of the hybrid-n parameter at Ie
~=
50 mAN
rbb'
ree
80 kn
Ce = 3 pf
=
=
100 n
=
1.3 rnA are as follows:
rb'e = 1 kn
Ce
=
100 pf
rb'e = 4 Mn
These values depend upon:
1. Temperature
2. Value of Ie
92
Electronic Circuit Analysis
3.3
Determination of Hybrid-x Conductances
3.3.1
Transconductance or Mutual Conductance (gm)
+-Ic
B
E
PNP
--------~~----------~--------~
Fig. 3.3 PNP transistor amplifier
The above figure shows PNP transistor amplifier in Common Emitter configuration for AC purpose,
Collector is shorted to Emitter.
..... (1)
leo opposes IE' IE is negative. Hence Ie = leo - a o IE a o is the normal value of a at room
temperature.
In the hybrid - 1t equivalent circuit, the short circuit current =
~
Vb' e
Here only transistor is considered, and other circuit elements like resistors, capacitors etc,
are not considered.
I
ale
gm = aVb'e VeE; K
Differentiate (l) with respect to Vb' e partially. leo is constant.
gm = 0 - a o OlE for a PNP transistor, Vb'e = -VE
oVb'e
Since, for PNP transistor, base is n-type. So negative voltage is given.
gm
=
a
0
OlE
oV
E
OV
If the emitter diode resistance is ,re, then re = u:::II E
E
But for a diode,
1l,VT
r=-I
':I=Io(ev/!]VT-l) I:::Jo·ev/!]vT
High Frequency Transistor Circuits
93
11.VT
r= -1- 11 = 1,
Neglecting Ico'
~ =
(Transconductance or Mutual Conductance)
~ is directly proportiortal to Ic. ~ is also a. -.!....
T
For PNP transistor, Ic is negative.
I~~ ~fl
will become positive. Since Ic is negative. So ~ is always positive.
V T is volt equivalent of temperature V T = Till ,600
At room temperature, T::: 300 K,
gm
=
IIel,
Ic is in rnA.
26
3.3.2
= 1.3 rnA,
~
= 0.05 AN
If
Ic
If
Ic = 10 rnA, gm = 400 mAN
Input Conductance (gb')
At low frequencies, capacitive reactance will be very large and can be considered as Open circuit. So
in the hybrid-1t equivalent circuit which is valid at low frequencies, all the capacitances can be neglected.
The equivalent circuit is as shown in Fig. 3.4.
r---T---~~-OC
E~------
____
- L_ _~_ _ _ _ _ _ _ _ _ _~_ _~_ _~E
Fig. 3.4 Equivalent circuit at low frequencies
Elect~onic
94
Circuit Analysis
The value ofrb'e» rb'e (Since Collector Base junction is Reverse Biased)
So Ib flows into rb'e only. [This is lb' (IE - I~ will go to collector junction]
..
Vb'e:::lb·rb'e
The short circuit collector current,
Ie =~. Vb'e;
Ie =~. lb' rb'e
hfe =
~cl
Vb'e = lb' rb'e
= gm' rb'e
B VCE
or
But
3.3.3
Feedback Conductance (gb' c)
hre = reverse voltage gain, with input open or Ib = 0
Vb'e
Input voltage
= - - = -.!........--:::Vee
Output voltage
rb'e
~=-~-
rb'e +rb'c
[With input open, i.e., Ib- 0, Vee is output. So it will get divided between rb'e and rb'e only]
or
hre (rb'e + rb'e) = rb'e
But
rb'e [1 - hre] = hre rb'e
hre « 1
rb'e
rb'e= - h
re
or
~ =
10-4
High Frequency Transistor Circuits
3.3.4
9S
Base Spreading Resistance (r bb')
The input resistance with the output shorted is hie. If output is shorted, i.e., Collector and Emitter are
joined, rb'e is in parallel with rb,c.
But we have seen that rb'e = hre · rb,c
hoe in very small and rb,c » rb'e
rb'e is parallel with rb,c is only rb'e (lower value)
hie
or
=
rbb, + rb'e
Irbb, = hie hie
=
rb'e
I
rbb , + rb'e
But we know that
rbb, is small, few Os, to few hundred Os
3.3.5
Output Conductance (gee)
This is the conductance with input open circuited. In h-parameters it is represented as hoe.
For Ib = 0, we have,
But
Ve
Vce
_c_+
+ ~ Vb'e
rce rb,c + rb'e
Ie
=
h
=-
re
Ie =
Vb'e
Vce
Vce
rce
--+
Vb'e = hre· Vce
Vce
+ ~. hreO Vce
rb,c + rb'e
96
Electronic Circuit Analysis
But
IC
h
=-
Vce
oe
So dividing by Vee and that rb'e »rb'e' rb'e + rb'e ::: rb'e
hoe
1
= -
rce
1
+ - - + ~ . hre
rb'c
= gee + gb'e + gm h re
But
or
If
gee = hoe - (1 + h fe) . gb'e
h fe » 1, 1 + h fe ~ h fe
I
gee
= hoe -
h Ce • gb'e
I
But
gb'e = h re . gb'e
But
gee = hoe - h fe . h re . gb'e
h fe . gb'e = gm
gee=hoe-~·hre
3.3.6
Hybrid - 1t Capacitances
In the hybrid - 1t equivalent circuit, there are two capacitances, the capacitance between the CollectorBase junction is the C c or Cb'e' This is measured with input open i.e., IE = 0, and is specified by the
manufacturers as COb" 0 indicates that input is open. Collector junction is reverse biased.
where
1
n ="2 for abrupt junction
=
113 for graded junction.
High Frequency Transistor (:ircuits
97
Ce = Emitter diffusion capacitance COe + Emitter junction capacitance CTe
C T = Transition capacitance.
Co =
Diffusion capacitance.
COe » CTe
Ce~ COe
C Oe a. IE and is independent of Temperature T.
3.3.7
The Diffusion Capacitance
For pnp transistor, base is n-type and emitter is p-type. So ifE-B junction is forward biased, holes are
injected into the base. The distribution of these injected holes between E and C is as shown.
The collector is reverse biased. So the injected charge concentration p' at the collector junction
is zero since they are attracted because of the negative potential at the collector. The base width W is
assumed to be small compared to the diffusion length Ls of the minority carriers. If W << Ls, then
p' varies almost linearly from the value p' (0) at the emitter to zero at the collector. The stored
charge in the base is the average concentration p'(O) times the volume of the base W A
2
Emitter
Collector
x=o
X=W
Fig. 3.5 Variation of charge in the base region
(where A is Cross Sectional Area), times charge Q.
W.A = Volume;
W = Base width ;
A
= cross sectional area of the juaction.
Q
=
Charge
p'(O)
-2-
Diffusion current
I
=
.
Average concentratIOn.
dp'
= - AQ . DB' dx = - A Q DB
DB
=
,
P (0)
p'(O)
W
diffusion constant for minority carriers in the base
=
I.W
AQDB
Electronic Circuit Analysis
98
·
d'f~'
.
CDe
Emitter
I lusion capacItance
=
dQB
dV
[E~[+~~T J
But
2
W .1 E
CDe = 2DB V
T
W2
= gm' 2DB
In the above derivation, we have neglected the base width i.e, there is no possibility for
recombination in the base hence Ib = O. Then IE = Ic' But actually it cannot be so. Hence, the
hybrid-n model is valid only when change in VBE is very small and so change in IE is equal to change
in Ic' i.e., Hybrid-trmodel is valid only under dynamic conditions, wherein change in IB is negligible.
So change in IE is equal to change in Ic'
Giacoletto who proposed the hybrid - n model, has shown that the hybrid parameters are
independent of frequency when
W2
2nf. --- « 1
6D B
Where
W
=
..... (1)
Base width
DB = Diffusion constant for minority carriers in the base
f
But
Frequency of the input signal
W2
Ce=~'W
B
2
C = __
W
_e
gm
2DB
W2
C
_,_=_e_
6D B 3g m
..... (2)
High Frequency Transistor Circuits
99
e -~
Ce =_1_
21t iT' gm
21tfT
e-
W2
1
--=-60 B
61tfT
21tf
, 61tfT
.. Equation becomes - - «1
or
f« 3 fr
:. Hybrid "model is valid for frequencies upto::: f13.
3.4
Variation of Hybrid Parameters with IIeI, IVeEI and T
3.4.1
Transconductance Amplifier or Mutual Conductance (gm)
W
VT
gm is ex: Ic
Vr = T/ll,600
1
~ex:T
gm is independent of VCE
Since in the active region of the transconductance, Ic is independent of VCE'
3.4.2
Base Emitter Resistance (rb'e)
hfe,VT
Ie
1
Ie
r
3.4.3
I
be
increases as T increases since fb' ex: Vr
e
Emitter Capacitance (C.)
As VCe increases, W the effective base width decreases
Ce decreases with
VCE
(increasing)
100
3.4.4
Electronic Circuit Analysis
Collector Capacitance (C c)
Cc depends on VCE
Ce oc (V CEt"
Cc~independentoflc
Cc is independent ofT
Cc decreases with increase in V CE
3.4.5
Base Spread Resistance (rbb,)
rbb, decreases with increase in IeSince as Ic increases, conductivity increases. So rbb' decreases, because of conductivity
modulation. But rbb' increases with increase in Temperature. Because as T increases, mobility of the
carriers decreases. So conductivity decreases. So rbb, increases.
3.4.6
Collector Emitter Short Circuit Current Gain
Consider a single stage Common Emitter. Transistor amplifier circuit. The hybrid-1t equivalent circuit
is as shown:
gb'c
rbb'
B'
C
+
+
Cc
gee
gb'e
Ce
E
3.6 Hybrid - 1t equivalent circuit in C E configuration
If the output is shorted i.e. RL = 0, what will be the flow response of this circuit? when
RL = 0, Vo = 0. Hence Av = 0. So the gain that we consider here is the current gain IL I Ic '
The simplified equivalent circuit with output shorted is,
I,
B'
C
Ce +C c
E
E
Fig. 3.7 Simplified equivalent circuit
A current source gives sinusoidal current Ic' Output current or load current is IL· gb'c is
neglected since gb'e « gb'e' gee is in shunt with short circuit R = 0. Therefore gee disappears.
The current is delivered to the output directly through Ce and gb'c is also neglected since this will be
very small.
High Frequency Transistor Circuits
101
I·1
_
Vb'e -
.
, gb'e + Jro
(Ce + C)
c
AI under short circuit ccnclition is,
A.
But
IL
--=
-
Ii
-gm
gb'e + jro(C e +C c )
- gm C +C C
h ' e
c::: e
fe
gb'e -
-h fe
A
But,
-h
=
fe
l+i(:']
Electronic Circuit Analysis
102
Where
At
/i3 =
h fe 21t(ce + CJ
Ai
J2
1
=
=
0.707 ofhfe ·
The frequency range upto f~ is referred to as the Bandwidth of the circuit.
Note: Common Emitter short circuit current gain was skipped out.
3.5
The Parameters IT
fT is the frequency at which the short circuit Common Emitter current gain becomes unity.
i.e.,
Let at
or
Ai
1,
=
I"
f -- JT'
ho
A.I
1+
hr. :: ~
1fT ::: hfe
J;
But
=
p
fT
=
1+(~;
1
=
=
when Ai ~ 1
gm
h fe {C e + C c }
fp . hfe =
k 2!~e I
::
=
1
r
(~; r (~; r
.Jp I
Ce » Cc
T
r
~ 1+(~;
~fu)2 ~
But
hfe
or
gm
21t(C + C )
c
e
High Frequency Transistor Circuits
103
~ 400 MHzs
::r::
'-'
i
200 Hzs
100 Hzs
10
10
100
~
Ic(mA)
Fig. 3.8 Variation of Ie with frequency
fT is very important parameter of the transistor. This indicates the high frequency performance
of the transistor. fT depends upon the operating point of the transistor. The variations of fT with Ie
is as shown,
fT = hfe.Jp
hfe is the short circuit current gain.
fp
is the frequency.
ThereforefT is the short circuit current gain Bandwidth product. So if we have two transistors
with same fT' the transistor with low hfe will have larger Bandwidth and vice versa.
3.5.1
Measurement of IT
The value of fT is very large of the order of few hundred MHz. To determine the fr of a given
transistor, we need eRO whose frequency range is of that order. Such a eRO is not common.
So how to determine fT with a practically available eRO which can respond upto any few MHz?
2
A=
I
hr.
H~n
1/2;
(f. J
»1
p
m~fP;re
IAII f ::::fp . hfe
=
fT
If we determine Ai of the transistor at a frequency fwhich is 3 or 4 timesfp but much smaller
thanfI' the product ofAi andfgives the value offr
For a transistor withfT = 80 MHz andfp = 1.6 MHzs, the frequency fat which Ai is determined
can be 5 times fp ~ 8.0 MHz maximum. We can have eRO which can respond at 8 MHz but not
80 MHzs. ThereforefT can be easily determined by measuringAj atf= 8 MHz.
104
Electronic Circuit Analysis
Expression for If»
3.6
-gm
Dividing by gb'e' Numerator and Denominator,
-gm igb'e
A = j2n f(C +C )
1+
e
c
gb'e
we know that
gm
gb'e=
h
fe
gm =h
gb'e
fe
-h ti
A -
e
- 1+
jf[_21t~-,----Ce_+_Cc-"'-'-)l
gb'e
But we know that A. =
I
-h fe
f
1 .
+J-
fp
Corrtparing,
B'
ii
Vi Vb'.
!1
gb'e
C
e
E
Fig. 3.9 Equivalent circuit
High Frequency Transistor Circuits
3.7
105
Current Gain with Resistance Load :
fr == ~.
hr. == 21t
(C e + CJ
Considering the load resistance Ru
Vb'e is the input voltage and is equal to VI
Vce is the output voltage and is equal to V2
Vce
~==-
Vb'e
This circuit is still complicated for analysis.
Because, there are two time constants associated with the input and the other associated with
the output. The output time constant will be much smaller than the input time constant. So it can
be neglected.
K == Voltage gain. It will be » 1
gb'c (K; 1)
:::: gb'c
gb'c < gce
rb'c :::: 4 Mn,
So gb'c can be neglected in the equivalent circuit.
rce == 80 K (typical values)
In a wide band amplifier RL will not exceed 2Kn, sincefH oc
RL
IfRL is smallfH is large.
fH == 21tC s (RC IIR )
L
Therefore gce can be neglected compared with RL
Therefore the output circuit consists of current generator gm Vb'e feeding the load RL so the
Circuit simplifies as shown in Fig. 3.10.
I
I
B'
r-------~----~--_r.c
Cc(l+gm-'L
R )
Fig. 3.10 Simplified equivalent circuit
Vce
K == -V == - gm RL ;
b'e
gm == 50 rnA\ V,
RL == 2Kn (typical values)
K==-lOO
So the maximum value is gb'c (1 - K) :::: 0.02595. So this can be neglected compared to
gb'e:::: 1 mAN.
Electronic Circuit Analysis
106
RL should not exceed 2KQ, therefore ifRL > 2KQ, Cc (1 + gm R L) becomes very large and so
band pass becomes very small.
cc= [K;I]
when
RL = 2KQ,
the output time constant is,
RL . Cc = 2
x
~Cc
103
x
3
x
10-12 = 6
x
10-9 S = 6 Il.sec. (typical values)
Input time constant is,
[Ce + C c [I + ~ Rd] = 403 Il.sec. (typical values)
So the band pass of the amplifier will be determined by the time constant of the input circuit.
rb'.
The 3db frequency fM = 2
gb'e
C = -1trb'e
21tC
C = Ce + Cc (1 + gm RL)
where
3.8
1
Miller's Theorem
It states that if an impedance Z is connected between the input and output terminals, of a network,
between which there is voltage gain, K, the same effect can be had by removing Z and connecting an
Z
ZK
impedance Zj at the input = (I _ K) , and Zo across the output = (K -I) .
jC
Vo
E~_ _~_ _~________~_ _~~~t
Fig. 3.11 High frequency equivalent circuit with resistive load RL
Cb'c = C c
rb,c and C c are between the input termed B' and output termed C. The voltage gain of the
V
amplifier = ~ = K (> > I). Therefore by Miller's theorem, Cc and rb,c can be connected between
Vb'e
C
B' and E (input side) with values = 1-~ and rb,c (1 - K) respectively. On the output side between
collector and emitter as
(K -1)
Cc-K
rb'c.K
.
and -(- ) restmg .
K-l
High Frequency Transistor Circuits
107
Therefore high frequency equivalent circuit using Miller's theorem reduces to, (neglecting rbb , )
Vce
K=Vb'e
Negative is used since current direction is opposite.
Vee = - Ic . RL ;
B' 0 - - - , - - - - , - - - - . . , . - - - - ,
.
~c
.----.----r---,----.--oc
~
I
::::,
C
."
.0
00
;}
bile
E o--~--~---~--~-----~~---~------~----~----~~E
Fig. 3.12 Circuit after applying Millers' Theorem
-IC·R L
K= --=--=Vb'e
But
IC
Vb'e
=
~
K=-gm' RL
3.9
CE Short Circuit Current Gain
This is the circuit of transistor amplifier in common emitter configuration.
Fig. 3.13 C E Amplifier circuit
The approximate equivalent circuit at high frequencies, with output shorted is,
~I,
H'
C
E
Fig. 3.14 Simplified equivalent circuit
Electronic Circuit Analysis
108
rb'e is assumed to be very large. So it is open circuit.
ree disappears since it is in shunt with short circuited output.
IL == - ~ Vb'e
Negative sign taking the direction of current into acount. IL is contributed by the current
soutce only.
1
Vb'e =IxZ =Ixy
Vb'e ==
IL
=
I·I xl
• C
gb'e + JCO e
-gm .I i
. C
gb'e + JCO e
Conductances in parallel get added.
Fig. 3.15 Conductances in parallel
Therefore current gain under short circuit conditions is,
A
=
i
~=
Ii
-gm
gb'e + jcoC e
But
-1
1+ jhfe
IT
·1
High Frequency Transistor Circuits
109
A=
1
when I
=
I~, Ai
falls by
"*'
or by 3db. The frequency range I~ is called Bandwidth of the
amplifiers.
/il:
Is the frequency at which the short circuit gain in common emitter
configuration falls by 3 db.
IT:
This is defined as the frequency at which the common emitter shunt
circuit current gain becomes 1.
-h
A.=
1
Let
fe
I+{~l
Aj = 1
1= IT'
hfe
r
+r
..
1=
00
1
I+(~;
(ITIp ~h'fe
r
(ITIp ~hfe'-I -_ hfe
..
liT = I~
0
2
hfel
/il is the Bandwidth ~f the transistor
hfe is the current gain
00
IT is the current gain, Bandwidth product.
/"
hfe » 1.
Electronic Circuit Analysis
In Common Emitter configurations, Ai >> 1. But as frequency increases Ai falls. Why should
Ai decrease? Ai
I
As frequency increases, Xe increases. So, Xe increases. i.e., more and more
Ii
number of carriers will be stored in the base region itself. Due to this, less number of carriers will
reach collector. Storage of carriers at the base and emitter increases. So Ie decreases. Therefore IL
decreases i.e.,.~ decreases. So ~ V b'e decreases or IL decreases. Hence Ai decreases.
IT depends on the operating point of the transistor. The graph of IT V 5 Ie for a transistor is
as shown,
= .-b. .
i
e;
i
400 MHzs
200 Hzs
100 Hzs
10
50
100
~
Ic(mA)
Fig. 3.16 Variation of Ic with frequency
For a typical transistor,
IT
= SO MHz
Ip
= 1.6 MHz
Example : 3.1
Given a Germanium PNP transistor whose base wiuth is lo-t cm. At room temperature and for a DC
Emitter current of 2mA, find,
(a) Emitter diffusion capacitance
IT
(b)
Solution:
Given
DB =47
CT is negligible,
Ce :::CDe
W2
But
:::
~-
~
I Ic I
= -V
2DB
Ie::: IE
T
IE
W2
C =-x-e
VT 2DB
2x10-3 xIO-S
26xl0-3 x2x47
= S.2 pf.
High Frequency ·Transistor Circuits
111
VT ·21tC e
=
1500 MHz
Example : 3.2
Giv~n the .follo~ing tr~sistor~easurerne)'1lS mad~at Ic = 5 rnA ; VCE = 10V and at room temperature
hfe
-
100 , ~e - 600n , [Aiel
10 at-lo1vtHz Cc - 3 pf.
-
Find f~, fT' Ce, rb'e and rbb ,·
Solution:
hfe = 100,
Aie =
10, atf= 10 MHz
FW
100
10=
1+
[L)
2
=
f~
[~)
100xtOO
10xlO
=
100
2
= 100-1 =99
2
[{) = 99,
I:. f~
=
f= 10 MHz for use,
1.005 MHz
A =hfe · f~ =100 x
C
II
I
1.005 MHz
=~.
~=1l
21tfT'
V
T
=100.5 MHz
Electronic Circuit Analysis
112
IIel
C =
V T .21t f T
e
Ce
5xl0-6
26 21tx 100.5
=
hfe
rb'e = gm
=
304 pf
100
= 5/26
= 5200
rbb' = hie - rb'e = 600 - 520 = 800
Example : 3.3
A single stage Common Emitter amplifier is measured to have a voltage-gain bandwidthfH of 5 MHzs
with RL = 500 n. Assume hfe = 100, ~ = 100 mAN, rbb' = 1000, Cc = 1 pf, andfT = 400 MHzs.
Find the value of the source resistance that will give the required bandwidth.
Solution:
gm
100 x 10-3
C - --= 0.0405 x 10-9 Farads
e - 21tfT - 628x400xl06
h
= -
lib'
fe
g
e
C =
=
lKn
m
Ce + Cc (1 + gm R L)
= 40 + 1 (l + 100
Bandwidth :::: fH sinee
It
X
10-3 x
1
"2
x 103 )
= 91 pf
is very small
1
fH = 21tRC
where
R = R;" rb'e and R; == Rs + rbb'
R==
1
21tfH'C
0.35 x
=
R~xrb'e
,
Rs +rb'e
Rs'
Rs
=
1
=-----:;----..-:;12
6
103
6.28x5xl0 x91xl0-
== 3500
= 3500'
'
rb'e == IKO
5390
= RSI - rbb'
== 539 - 100 == 4390
113
High Frequency Transistor Circuits
Example : 3.4
Show that at low frequencies, the hybrid
approximate CE h-parameter model.
~~
1t -
model with rb'e and ree taken as infinite reduces to the
+--Ic
B
C
hIe
Eo---------~----------+_--------------_oE
Fig. 3.17 Circuit diagram for EX.3.4
Solution:
The h-parameters equivalent circuit in Common Emitter configuration is as shown in
Fig. 1.49. We have to show that the hybrid 1t equivalent circuit will also be the same with the
approximation given in the problem.
E----------~---------+---------------oE
Fig. 3.18 Simplified circuit
If the expressions for the output current I and input impedance hie are same in the hybrid 1t
equivalent circuit and h-parameter equivalent circuit, we can say that both the circuits are identical.
This is the hybrid - 1t equivalent circuit neglecting capacitances
Ii,
be
=
ree =
00
(since at low f, they are (0).
1= gm' Vb'e;
Vb'e = lb' rb'e
I
we know that
=
gm' lb' rb'e
rb'e gm = hfe
1= h fe · Ib
[in the derivation of gb'e]
[which is the same as in h-parameters circuit]
hie = rbb' + rb'e
Example : 3.5
The following low frequency parameters are known for a given transistors at Ie = lOrnA, Vce = 10v
and at room temperature hie = soon, hoe = 10-5 AN, h fe = 100, h re = 10-4U .
At the same operating point, fT = 50 MHz, and Cob = 3 pf, compute the values of all the
hybrid - 1t parameters.
Electronic Circuit Analysis
114
Solution:
IIcl
lOrnA
g = - = - - =385rnAN
m
VT
26mV
h fe
100
rb'e = - = - - = 260n
gm
0.385
rbb' = hie - rb,!; = 500 - 260 = 240n
rb'e
r,
= -h
= 260 x 104 = 2.6 Mn
bc
re
~ = hoe - (l + hfe ) gb'c
=4
x
10-5 -
101
2.6x106
= 0.120
x
106
I
r = =833Kn
ce
gce
gm
385xlO-3
Ce = 21t/T = 21tx50x106
=
1224 pf
Cc = 3 pf
Example : 3.6
A single stage Common Emitter amplifier is measured to have a voltage gain bandwidth product/H of
5 MHzs, with RL = 500n. Assume hfe = 100, gm = 100 mAN, rbb' = lOOn, Cc = 1 pf and
IT = 400 MHzs.
.
(a)
Find the value ofRs that will give the required Bandwidth.
(b)
With the value ofRs ' determined in part (a), find the midband voltage gain VrfYs'
Solution:
gm
100x10-3
C =
=
= 0.0405
6.28x400x 108
e 21t1 T
x
10-9 Farads
hfe
rb'e= = IKn
gm
C = Ce + Cc (1 + ~ RL) = 40 + 1 (1 + 100 x 10-3 x
1
2"
x 103) = 91 pf
Let Rs' = total input resistance = Rs + rbb' ; Rs is the resistance of the source.
Let R be the equivalent input resistance = Rs I " rb'e
IH = Higher cut off frequency
1
21tRC
=--
High Frequency Transistor CIraIIts
R=
1
21t f H ·C
115
=
1
2x1tx5x10 6 x91xlO- 12
= 0.35
x
103 = 3500
,
1
gmRvGs
Av in the midband region = G' +g ; GS=RJ
S
b'e
s
Avs =-32.5
Redraw the Common Emitter hybrid-1t equivalent circuit with the base as the common terminal
and the output terminals, collectors and base short circuited.
Common Emitter Hybrid-1r Equivalent Circuit
rb'c
C
\r-----~~----~----~c
B
r ce
E _________....L..__- . L_ _ _ _ _ _ _ _ _ _"T"--.L_ _ _ _ _ _ _ _..L..-_ _ _ _ _ E
GND
Fig. 3.19 CE Hybrid - 1t equivalent circuit
The base should be common terminal. So draw a line between base B and B'. We have rbb" So draw
another line to indicate B'. Between B' and E, we have Cb'e parallel with rb'e' So draw these two in
parallel and indicate point E. E and B are input points. C and B are output points. Between B' and C,
WE' have Cb'e and rb,c' So draw these two in par~llel and indicate the point at Collector. Between
r ce
E
c
----..,.-----1
r ce
w----L-------r---------~----w
B~----------------~r----------------------+--~
GND
Fig. 3.20 With rbb , terminals 8' and 8 shown separately
B
Electronic Circuit Analysis
116
Collector and Emitter we have ~ V b'e in parallel with ree' So indicate these two between Emitter and
Collector output terminals. Base and Collector are to be shorted. So short Collector and Base terminals.
Therefore the voltage drop across the resistor
=R
x
I = 24 Kn
x
100 rnA
= 2400 volts!
A very large value
DC powers dissipation = Ioe2 R = (100
=
240000
, =
240 W
x
x
10-3)2
x
24
x
103
10-3
Such a resistor cannot be easily obtained. Hence replacing L by equivalent R can be done only
when the output current is small.
3.10
Hybrid - 1t (pi) Parameters
To analyse the behaviour of transistor at high frequencies, a model based upon transmission line
equation will be accurate. But the resulting circuit becomes very complicated. By making
approximations, it becomes simple, but the analysis may not be accurate. So hybrid-x model is a
compromise between the two.
3.10.1 Hybrid -1t Common Emitter.Model
Common Emitter model which is valid at high frequencies is called hybrid-x or Giacoletto model. It is
as shown in Fig. 3.21.
B
~~--~--------~~~~--~------~--------~----~c
E----------~--------~--~------------~--------~-----E
Fig. 3.21 Hybrid - x - C E Model
Ce = Diffusion Capacitance or Emitter Capacitance.
rbb, = Base s.preading Resistance
~
Vb,e = current source
gb'e = conductance
Ce = collector junction barrier capacitance.
By analyzing the circuit we get results which will agree with practical results at all frequencies.
All the hybrid - 7r parameters are assumed to be independent offrequency.
High Frequency Transistor Circuits
117
3.10.2 Discussion
Node B' is not accessible rbb , is base spreading resistance. Excess-minority carrier concentration
injected into the base is proportional to Vb'e' So the small signal collector current with the collector
shorted to the emitter is proportional to Vb'e' This effect accounts for the current generator ~Vb'e'
The increase in minority carriers in the base results in increased base current. This is taken into
account by the conductance gb'e between B' and E. The excess minority carriers stored in the base is
accounted for by the diffusion capacitance Ce' The feedback effect between input and output is taken
into account by conductance gb'e' Ce is collector junction barrier capacitance. Typical values of
hybrid - 1t parameters are for Ie = 1.3 rnA at room temperature :
~
= 50 mA/V, rbb' = 1000,
rb'e = 1KO, Ce = 100 pf,
Ce = 3 pf,
ree = 80 KO
3.10.3 Transconductance: gm
leo
=
Reverse saturation current.
Figure shows PNP transistor in CE configuration. Collector current Ie = leo -
0.0
IE'
~I
B'
Tbb'
B
VB'E
=
+v
t
Cl o
-
ce
E--------~~------~--~----------~
Fig. 3.22 PNP transistor circuit in CE configuration
olc
Transconductance gm = oV ,
I
BE VcpK
=
OlE
0. 0 -;--V =
v B'E
+
0. 0
al C
aVE
for PNP transistor.
VE = VB'E as shown in the above Fig. 3.22.
If the emitter diode resistance is re, then re = 0 VEl
a IE
VT = voltage equivalent of temperature
Since
~,IE
depend on temperature. For PNP transistor, Ic is negative, for NPN Ie is positive.
3.10.4 ~ Input Conductance gb'e
Fig. 3.23 (a) shows hybrid-1t model valid at low frequency. Neglecting all the capacitances,
Fig. 3.23 (b) shows the same using h-parameter equivalent circuit.
118
Electronic Circuit Analysis
~Ic
B
\~---T---'----~--~C
r ce
Eo-------~~--~--------~~--~------_oE
(a) Equivalent circuit
~
+--1
______~~c______~
C
hoe
E ~------------~--------~------~-------OE
(b) Simplified circuit
Fig. 3.23
Short circuit collector current Ic = ~ Vb'e - ~ Ib rb'e
Short circuit current gain hCe
or
rb'e
=
= ;c
hfe
gm
b
Iv
CE
=
hfe VT
= lIe I
~ rb'e
or
gb'e
gm
=-
hCe
rb'e is directly proportional to temperature and inversely proportional to current.
3.10.5 Feedback Conductance: gb'c
With the input open circuited, h re is defined as the reverse voltage gain. From Fig. 3.24(b),-with
Ib = 0,
or
rb'e (1 - hre ) = hre rb,c
But
rb'e «1, rb'e:': ~e rb'c
or
gb'c = hre gb'e
Identification of leads :
Transistor when held with leads upwards E is on left half side. The upper half semicircle is emitter.
E B C
6
Transistor case
A
-=--<. . . _---'1-=- K
A __
~"""I-K
Diode lead identifiation
Fig. 3.24 BJT and diode devices
High Frequency Transistor Circuits
119
I
h. = Input Z = Ve output shorted
Ie
Ie
hoe = output admittance
~I
input open
Vc
hre = reverse voltage d.c gain = Vb'e Vee
hfe = forward current gain =
~I
Vee = k
IB
3.10.6 Base Spreading Resistance r bb,
The input resistance with the output shorted is hie' Under these conditions, rb'e is in parallel with rb'c'
But
gb'e
=
hre gb'e
Since rb'e»
rb'e
Therefore Parallel combination
II rb'e - rb'e
rb'e
3.10.7 Output Conductance gee
With the input open circuited, the conductance is defined as hoe for Ib = 0,
v.
V
Ie = ~ +
ce
+ ~ Vb'c
rce rb'c + rb'c
~
But
= Vb'e 1Vee
hoe =
:. Vb'e =hre Vee
Ic
1
1
- = - + - +~hre
Vc rce rb'c
By substituting the equation gb'e
=
~m
, gb'c = hre gb'e above,
fe
or
If
~
= hoe -
hfe »
(l + hfe ) gb'e'
1, {l + bee> ::: hfe
gee = hoe - bee gb'c
But
gb'e
= ~.
gb'c
~e = hoe - hfe hre' gb'e
But
bee' gb'e
=
~
Electronic Circuit Analysis
120
3.10.8 Hybrid - 7t Capacitances
The hybrid Ce
:
model consists of two capacitances Cc and Ceo
The collector junction capacitance C c = Cb'c. It is usually specified as Cob by the
manufacturers. In the active region, collector junction is reverse biased, C c is the transition
1t
capacitance and varies as (VcEtn where TI is
C e:
1
2"
1
or
"3' for abrupt or graded junction.
It represents the sum of emitter diffusion capacitance C De and emitter junction capacitance
C Te . For a forward biased emitter junction C De is usually much larger than C Te,
CDe + C Te ::: CDe'
Stored charge in the base QB =
I.W 2
2D
B
DB = Diffusion constant for minority carriers in the base
W = Base width
= Diffusion current
Emitter diffusion capacitance CDe =
dQ
d:
=
rate of change of Q B with V.
or
dV
where
re
V
= dI = -I-T
emitter junction incrematal resistance.
E
C De =
2
W IE
2D B ,VT
---=W2
gm = 2DB
IT:
Therefore Diffusion capacitance is proportional to emitter bias current IE'
Hybrid - 1t models are valid for frequencies upto l'l:i 113.
It is defined as the frequency at which the common emitter short circuit current gain
becomes 1.
h fe = short circuit current gain
The graph of IT vs (versus) Ic is as shown. It depends V CE = 5V upon operating conditions
121
High Frequency Transistor Circuits
~ 400 MHzs
::t
'-"
~
200 Hzs
i
100 Hzs
10
50
100
- 7 Ic(mA)
Fig. 3.25 fT Vs Ie graph
3.11
Measurement of
IT
IT is very high, to measure by experiments
IAil =
for I»
~,
hfe
-======
Ai = current amplification
FW
1 can be neglected
IAII . I = hfe~ or
..
·1 = hfe/l3
IAil
= IT
I/T=hfe.f131
At some frequency 1;, measure gain IAJ Then
IT = 1;
Ex:
IfIT = 80 MHz,
~
IAil
= 1.6 MHz,
1; = 5 x 1.6 = 8 MHz, which is convinient to measure than 80 MHz,
At
1;
=
8 MHz, AI
IT = 1;
=
10
(All) = 80
IT can be used to measure Ce
Variation of Hybrid-x Parameter with Voltage, Current and Temperature
3.12
1.
qlIc I
gm = ~. As Ie increases ~ various linearly. It is independent of VCE' It varies
inversely with temperature, Incremental transconductance of a transistor.
Electronic Circuit Analysis
122
2.
Cc
3.
PO:
:
VCE-n
Low frequency value of P
Po = gm rb'e
Independent of Ic'
Increases steadily with VCE
Increases with temperature T.
rlt
=
Incrematal resistance in hybrid
3.12.1 Determination of Hybrid -
7t
-7t
model.
Parameters from Data Sheets
Suppose we want to determine the hybrid - 7t parameters for a 2N 3564 NPN silicon transistor at the
operating point of Ic = 5 rnA and VCE = 5V. Make reasonable assumptions.
Table 3.1 Characteristics of 2N 3564 NPN silicon transistor at T = 25°C.
Symbol
Parameter
Min.
Max.
Units
Test conditions
hre
Low frequency current
gain at/= 1KHz
20
80
-
Ic = 15 rnA, V CE = 10V
hre
High frequency current
gain/= 100 MHz
4
7.5
-
Ic = 15 rnA, VCE = 10V
Real part of hie
open circuit output
-
30
0
Ic = 15 rnA, VCE = 10V
2.5
pf
VCB = 10V, IE = 0
r b,
Cobo
Capacitance
At the desired operating point of 5mA,
Trans Conductance
=
0.2 mhos (U).
_=l=_
Vr
KT
25
rlt = Incremental resistance in the hybrid -
7t
model = rb'e'
hfe
80
r = = - =4000
b'e g
0.2
m
From the Table 3.1 Cob' open circuit output capacitance = 2.5 pf at VCB = 10V. (C c) = CIl will
also be the same at 10V. Therefore to calculate Cc at the desired operating point of 5V, we have
Cc = CJ.1 ~ (VCB)-1/3
n=
1
'3 for junction diode.
High Frequency Transistor Circuits
123
--------------------------------------------
To find C e ' (C 1t), incremantal capacitance in hybrid - 7t model.
7tJT = ro[hfe (ro)]
JT = J(h fe )
h fe = 7.5 at J= 100 MHz
roT
=
2
JT =7.5
'.
C
lOO =750MHzatI c =15mA
x
'g
e
=i
---.llL W
T
0
c
Ce =C~
C1t
=
gm
W
T
= ( o.
-Cc
7~0~2 7t) - 2.5
q
gm = KT x Ie rnA
Ie
=
= 126 pf at operating point of 10V
l=_
KT
25
15
~ = 25
15 rnA,
3
="5
= 0.6 mho (0).
0.6
C
1t
=
Ce =
27tx75Ox106 - 2.5
= 128.5 - 2.5
=
126 pf
To convert this to the desired operating point of Ie = SmA, since (C e varies linearly wh,1I c ).
5
C = 126 x e
15
=
42 pf
Electronic Circuit Analysis
124
Example : 3.7
Find wherever possible, appropriate values for the hybrid - 1t parameters at Ic = 5 mA,
V CE == 4V for a 2N 1613 transistor (BJT) using the data listed below. Use typical values. Make
reasonable approximations.
= 25°C
Cha!ac1eristics of 2N 1613 at T
Symbol
~
Characteristics
Minimum
Typical
Units
Maximum
Low frequency
Ihfel
f= IKMHz gain
35
80
-
High frequency
3
4
-
-
and gainf= 20MHz
Cob
output capacitance
-
18
25
pf
r b'
Real part of hie at
f= 350 MHz
-
30
-
n
Test Conditions
Ic=lOmA
V CE = 10V
Ic = 50 rnA
V CE = 10V
IE = 0
V CB = 10V
Ic = IO rnA
V CE = 10 V
Transconductance: gm
KT
-
at T = 25°C
=
25
5
25 = 0.2 mhos (U).
=
From low frequency h fe data we know that
Ic = 5 mA, Pwill be the same.
rlt
q
= -
Po
gm
80
=-
0.2
=
Po =
80 at Ic
=
I 0 mAo We can assume that at
400 ohms (n).
Incremantal resistance in hybrid - 1t model.
Feed back capacitance C Il = 2.5
IO)+j
(5
=
3.16 pf at VCB
=
5V :::..V CE'
80
.s;:.<I!.
60
i
20
40
10
50
100
~ Ic(mA)
Fig. 3.26 Variation of hfe with Ie
High Fri/quency Transistor Circuits
IT :
125
Frequency at which Current Gain
=
1
roT = ro Ih fe (ro)1
or
At
IT =
Ilhfe(f)1
1= 20 MHz,
h fe
=
4
IT = 4 x 20 = 80 MHz
Capacitance in Hybrid - 1t Model
Cn is in pf. If we take gm in Kn, C~ in pf, and roT in nano seconds.
0.2
--------;- - 3.16 x lO-12
2x7tx80xlO 6
lO6 xO.l _ 3.16 pf
3.14x80
=
105
251.20 - 3.16 pf
:::: lOO - 3.16 pf
= 96.84 pf.
Now we must convert it to the desired operating point of Ic = 5 rnA, V CE = 4V.
Cn::::C b
Cn varies Iinerly with Ie-
Ie
=
5 rnA, VCE '"" lOY,
= 96.84 x
to5
= 48.42 pf
To convert Cn to V CE = 4 V, instead of 10, we need to know the relation between
As it is not known we assume that Cn remains the same at VCE = 4V also.
rw at Ic = lO rnA, VCE = lOY
At
IC
=
4 rnA
=
30n
0)
and VCEO
126
3.13
Electronic Circuit Analysis
Specification of Amplifiers
The Specification parameters and ideal values of an amplifier circuit are, given below:
Parameters
1.
2.
3.
4.
5.
6.
7.
8.
Ideal Values
Bandwidth (BW)
Volgate gain (Av)
Current gain (AI)
Power gain (Ap)
Figure of merit (Gain - BW product)
Input Impedance
Output Impedance
Frequency Roll-off
00
00
00
00
00
d:;
0
00
Typical Values
100KHz
600
10
6000
1000KHz
5.6 Kn
900 n
2 db/decade; 8 db/octave
Example 3.8
If ~ = 150, what are the cutoff frequencies of th.e input and output lead networks of the
given circuit?
Rl = 10K
jr.5-~-;i,"F
--- :;.2K
R2
Fig. 3.27 CE amplifier circuit
When ~ value is given, and not hie of the transistor, the input impedance of the transistor can
be determined.
Yin =
Yin
== ~ i b· re'
Z in --
Rm
iere'
=
pre'
Pi b .re
l'
b
R1
=A.r,]t
P
~
II ~ 1/ P . re'
Kn
10K II 2.2K II 3.14 K = 1.l8Kn
= 150 x 22.7 = 3.14
Rm =
Ro = Rc = 3.6K
High Frequency Transistor Circuits
127
1
f.n
= 21t ( Rs + Rin ) C
m
Cutoff frequences of input network (HPF)
1
21t(Ro + RL )C o
10=-----
.!in = IH =
1
21t(lKO+1.l8KO)(0.47f.lF)
= 155 Hz.
fo
=
A=
21t(3.6K + 1.5 KO)(2.2f.lF)
= 14.2 Hz
14.2 Hz is LPF cutoff frequency of output network.
3.14 Design of High Frequency Amplifiers
Broad band transistor (BJT) amplifiers use shunt feedback and series feedback methods so as to get
large Gain - Band width (BW) product. If number of amplifier stages are cascaded, the output
current is made up from several transistors and the individual device operating point canbe made to
correspond to the optimem Gain - B.W. product of emitter current.
A single section of a multi stage amplifier is shown in Fig. 3.28. The input impedance of the
amplifier circuit is low compared to the impedance of the parallel R-C network.
-.,.--+vcc
R.F. coil
U2
U2
Fig. 3.28 One stage of multistage amplifier.
~
where ~ is the
~
1
=
21tRC
cut-off frequency of the transistor.
Electronic Circuit Analysis
128
Now consider 3 such stages cascaded as shown in Fig. 3.29. R is the series resistor and hie is the
transistor low frequency input impedance. hre of the BJT is small and so it can be neglected.
U2
U2
L
U2
Fig. 3.29 Three stage cascaded amplifier.
e 1 is the input voltage
.
Ibl=
el
el
(R+hie)-R
"-J
Since hie of the BJT is small compared to R.
RL is a 50
n load for the circuit.
The voltage gain gets multiplied for each stage. So for n-stages,
eo = (n.
ic) (~L )
=
n.
Po
ib (~L )
n. Po. e!. RL
2R
where Po is the low frequency current gain (P) of the BJT.
n = number of cascaded stages.
The Voltage Gain for 'm' cascaded stages is, (y'G) m
(V. G)m = [
no RL'
2R
po]m
Typically, a single stage will have a 3-db frequency of200 MHzs in high frequency amplifiers.
High Frequency Transistor Circuits
129
Effect of Emitter Degeneration :
The approximate high frequency equivalent circuit for a BJT operating in C.E. configuration is shown
Fig. 3.30
In
C
Collector
B
Base
•
/3ib Current source
Fig. 3.30 High frequency equivalent circuit for a 8JT.
rbb' = Base spread resistance
Cc = Collector emitter capacitance
I
CC2 = Collector-base capacitance
re
= Emitter resistance =
~~ =( ~~) Qat 25°C.
T = Temperature in OK; (273 + 25) = 298 oK for 25°C.
k = Boltzmans' constant.
Cb'e = Diffusion capacitance
Let
P be the current gain at any frequency f
Then
where Po = Low frequency current gain
f~ =
Beta cutoff frequency
Electronic Circuit Analysis
130
Typical values of these parameters are, for VCE
rbb'
=
=
IOV,
IE =
lOrnA
70 Q
CCI =
lpf
CC2 =
2 pf
Po = 100
fr, = 6 MHZs
re =4 Q
Cb'e = 130 pf
Iff»
fr"
the input circuit can be simplified as, (Fig. 3.31)
Base
Emitter
Fig. 3.31 Simplified input circuit.
The equivalent input impedance of the R-C combination on the input side along with transistor
capacitances is shown in Fig. 3.31.
If the frequency range of consideration is such that Xcel and Xce2 can be neglected and 'f' is
greater than f~,
f
> f~ but Xcel and Xce2 are negligible,
(P + l)Re
ZI:::: rbb' + (1 + jro ReCe)
But,
High Frequency Transistor Circuits
Upper cut-off frequency h
131
= ---
21tReCe
for frequencies below ii,
Z
~
+ Re ) + Re ~o
jro
1+rop
(tbb'
..o....::..:=----=.~---=.__=_
1-
with a parallel R-C network inserted in the base,
we ha\Te,
Z
=
T
] + ( rbb' + Re ) + ~ORe,
.R
[ 1+ JroCR
1 . ro
+Jcop
If the R-C combination has the same 3-dB points as/p' then/p
=
1
21tRC
so that
If f > Ip but less than h,
(/2
=
1 ), the input impedance of the delay line comprises a
21tReCe
resistance (rbb' + Re) in series with a parallel combination ofReq and Ceq'
Calculation of Re and C e for a required Gain - B.W value (for uncompensated case)
The voltage gain Av of one section at low frequencies is given by,
A
=
V
~o RL
--;::-----'---=----=-----=;2[(R+PoRe)+(rbb' +Re)]
2(R +~oRe)
For uncompensated case, Ce = O.
The Bandwidth B.W is given as,
where
1
C =
---,eq 21t/p (R + Po Re)
B.W:::
12
Electronic Circuit Analysis
132
f~. ~o RL
Gain Band width product = 2(
For compensated case, Ce
=F-
+
rbb'
R )
e
0
In order to have good compasation, Re Ce must be chosen such that the break frequency is same
as in the case of uncompensated Band width.
:. we have,
fiu
fi of uncompensated circuit,
=
fiu = 2n CeRe
The value offic' the upper cut-off frequency for compensated circuit can be obtained by equating
the real and imaginary parts of Zr at fie.
Since
fie > fW
We have
1+
Also,
ro Ce Re = 1
(jroJ ~ jro
ro~
ro~
Zr atf= fie.
~ -j [R +{{PoR, (I +iJ)1( :: )+rbb' +L~'j)]
~ [rbb' +(~'
)-CP::"R, )] - j [f (Po 2~:+2R) +(~' )]
p
. Equating real and Imaginary parts,
f~ (Ro + ~o Re)
fie =
Ii
bb'
The B.W improvement factor 'k' is,
k=f
f2c
2u
=1+(~J
rbb'
Lo and Co can be determined, using the transmission line equation,
Zo =
(Lo
ICo)[I-(
f
!cut-off
)2]
High Frequency Transistor Circuits
f
133
1
---==
cut-off - 21t~LoCo
C is usually chosen as 2 or 3 times Co
R is computed to correspond to R = 21t C/ •
13
Re can be calculated from the equation for fico
Ce is obtained from the equation for fiu'
Typical values are:
fcut-off = 250 MHz
Zo = 480
Lo = 0.05 IlH
Co = 19
pi
(R + ~o Re) = 2.5 kO
Re = 30 0
Ce = 50
pi
Problem 3.9 : Find Zj' Zo and Ay in the case of an emitter follower given that,
C be,
rb'e
=
=
1000 pf Cb'c = 10 pF
1000
rbb' = 3.0 0
R~ =
h fe =100
1000
RlI = 1900
Expression for midband input impedance is,
Zj (midband)
rbb + rb'e + (l + hfe ) Ret
= 30 + 100 + (l00 + 1) 100
=
= 130
+ (l01) (100)
Zj (midband) ::: 10kO
(rb' e + hfe Ret»> Ret
We have
C1 =
Cb'e
--"-"'-l+g mR e,
1000x 10- 12
(1 + 100)
::: 10pF
Electronic Circuit Analysis
134
(100+10 4 ) (10+10)10- 12
5
0)1 =
x
106 rad/sec
for emitter follower Ay < 1.
Problem 3.10: Design a single stage I.F. amplifier to have carrier frequency fe = 455 kHZs, B.W =
10 kHZs, Vee =
-
9V, Ie =
-
1rnA. The small signal hybrid 1t parameters are :
rbb
=
70
n
gb'e = 800 J..lV
gee
=
8.6 J..ln
Cb'e = 1550 pi
Cb'e = 9 pi
gm
=
38.6 mil
gb'e = 0.25 J..ln
Solution: The circuit diagram is shown below
To next stage
Fig. 3.32 I.F. amplifier single stage.
135
High Frequency Transistor Circuits
1
rb'e = g b'e = 1.25 kO
1
r == 120 kO
ce
gee
rb'e
gbb
(Or
1
g b'c
=
=
4 MO
0.133 mU
2rcir
=
=
Choose RI and R2 such that Ie
=
2.8 rad/sec
1rnA.
=
Since XCI «R I, ifRI = 5ill, than C I = 0.05 IlF
IfRI = 1kO, then C I = 0.11lF.
Let RL = 500 0 and CD = 0.05 IlF
Substituting in the expression for R, and Ro'
Ro = 32.4 ill.
R j = 526 0
C, = 1268 pF
Gm
=
Co = 33 pF
34.6 mAN.
I\ is calculated as 12.4 kO. C;y is calculated as 9 pF. Ro must be matched with the load (R, of the
next stage). So the transformer turns ratio must be,
_
n-
~Ro
3
32.4 x 10
526
=
Rj
= 7.8: I
The feedback components, resistor Rn and capacitor Cn can be determined as,
12.7kO
= 16400
7.8
Cn = n. Cy = 7.8 x 9 x 10- 12 = 70.2 pi
R
n
Ry)
= ( -n
=
The equivalent circuit is shown in Fig. 3.33.
c,
Fig. 3.33 Equivalent circuit.
136
Electronic Circuit Analysis
v2
_1_.
R '
p. =
1
R =R =n2 R
L
j
=
0
1
3
(34.8Xl0-
f.
3
32.4x10 x528
4
= 5200
Power gain in dB
=
10 log (5200) == 37.16 dB
Q factor =
fr
'B=
455 x 103
lOxl03 = 45.5
with a coil having Qc = 100,
The inductance of the coil is,
L=
Ro
(Qc -Q)
2cor Q.Qc
32.4 x 103 (1 00 - 45.5)
2 x 2.86x 106 x 45.5 x 100
= 67 IlH
Parallel tuning capacitance
1
~
2
( 8.18 x 1012 )
The total transformer primary inductance
•
1800pF
(68 x 10-6 )
~
is,
1
1
L = -=----::------2
T
cor c. (8.18 X1012
(206.2 ~ 10-12 )
r
LT = 590 IJ.H
137
High Frequency Transistor Circuits
Problem 3.11 : Design a JFET Single Tuned Narrow Band amplifier with a centre frequency of 5.0
MHZs, Q = 40, B. W = 100 KHZs midband gain Amid = 150. Given, for the JFET, CDG = 20 PF, CGS
= .sOPF, VDD = 15V, Vp = - 2V.
Solution:
Given
10= 5 MHZs
L=?
Q=40
C=?
Amid =
150
RD
=?
CDG = 20 pF
CGS = 50 pF
VDD = 15 V
Vp =-2V
~
= 10 mU
ro= 1 MW
Circuit Diagram :
+VDD
0.01 IlF
Fig. 3.34 Circuit diagram for Problem~3.11.
Electronic Circuit Analysis
138
Centre frequency
010 =
I
VL(C+C~D)
,
Quantity
R
facto~ Q = 010 Rp (C + CaD) = Olo~ ; Amid = gm' Rp
where Ro is the parallel resistance,
Rp = (ro
The maximum value ofRp =
II RD II R 3)
f '
O
10= 5 MHZs
010 = 21t 10 = 2 x 3.14 x 5 x
106 = 31.4 MHZs
1
..... (1)
40 =
R
p
..... (2)
Rp
..... (3)
31.4x 106 xL
AmId = gm'
150 = 10 x 10-3 x R p
ISO
Rp = 10 x 10-3 = 15kQ;
R p = 'likQ
High Frequency Transistor Circuits
L=
139
15 x 103
.
31.4 x 106 x 40
31.4 x 106
0.012
x
~0.012 x IO-\C + 20p!)
0.012xl0-3
= 0.012 mH
C
=
15kn
=
II R3 -o
= 83.3
10- 12 = 83.3 pF
x
:. RD =?
1011
106
II x 105 - 11
10- 12 = 10- 15
)
Rp = ro II RD I R3
IMOxlOOkn
(lMn+ 100kO)
x
(31.4 x 10
IMn
ro II R3
r
= 0.001
83.3 pF
Rs = 100 ill
=
2
6
10- 15
L
Rp
1
10-3 (C + 20 pF) =
= 83.3 - 20 = 63.3 pF
=
10-3 H = 0.012mH
1
C
ro
x
= ---;========
(c + 20 pF) =
Let
= 0.012
IOxl05 xl05
= (IOx10 5 + 105)
Electronic Circuit Analysis
140
15
x
109 = 103 RO (1000 - 165)
15 x 109 = 103 xRox 835
15 X 109
6
RO = 103 x835 = 0.012 x 10 = 12kQ
Ro = 12kQ
High Frequency Transistor Circuits
141
1.
Hybrid - 1t model is also known as model _ _ _ _ _ _ _ _ _ model.
2.
Hybrid - 1t circuit is so named because _ _ _ _ _ _ _ __
3.
b' to denote Base spread resistance rbb. is _ _ _ _ _ _ _ _ _ terminal of the transistor.
4.
Transconductance ~ in Hybrid - 1t model is defined as __________
5.
Typical value of rbb, is _ _ _ _ _ _ _ __
6.
Expression for gb'. = _ _ _ _ _ _ _ __
7. Expression for h,. in terms of hr. and Ie is, _ _ _ _ _ _ _ __
8.
fr is the frequency at which Common Emitter short circuit current gain _ _ _ _ _ _ _ __
9.
Relation betweenfr, hr.
and~
is _ _ _ _ _ _ _ __
10. Expression for C. interms ofgmandfr is, _ _ _ _ _ _ _ __
11. Typical value of Cob' output capacitance is _ _ _ _ _ _ _ __
12. rx in Hybrid -1t equivalent circuit is _ _ _ _ _ _ _ __
13. If hr. = 100, gm = 0.5 mhos, determine the value ofrb,o'
14. Hybrid 1t capacitance C" is of the order of _ _ _ _ _ _ _ __
15. Relation
betweenfr,~
and
lye is, _ _ _ _ _ _ _ __
16. Classify amplifiers depending on the position of the quiescent point of each amplifier.
17. Draw the hybrid -1t model for a transistor in CE configuration.
18. What isfr?
19. What is the significance of the gain bandwidth product?
20. What would you neglect while drawing a low frequency model?
21. How does the trans conductance
(~)
depend on current ?
22. How does the trans conductance (gm) depend on temperature?
23. Write the expression for rbe in terms of gm and h,.
Ie
24. How does the diffusion capacitance depend on current and temp?
25. Write the expression for CD. in terms of gm' W, DB'
26. When is the hybrid - 1t model valid? (at what frequencies)
27. What is an emitter follower?
28. Which time constant is considered for the bandwidth?
29. What is the expression for
30. How
are~
and fT related?
~
cutoff frequency ?
Electronic Circuit Analysis
142
1.
Draw the high frequency equivalent circuit of a BJT and explain the same.
2.
G:iYe file 1ypicalvaJuesofvar:iousHybri:l-1t parameters.
3.
Derive the expressions for Hybrid -1t parameters., Ce, rbb" rb'e' Cc
4.
Derive the expression for the Hybrid - 1t parameters gm' ree, Ce and rb'e, gee.
5.
Explain about Hybrid -1t capacitances. How do Hybrid -1t parameters vary with temperature?
6.
Obtain the expressions for lfi andlT of a transistor.
7.
Draw the circuit and derive the expression fOF'CE short circuit current gain Ai interms of at any
frequency 'f and ~ of the BJT.
8.
Explain how lfi and IT of a BJT can be determined? Obtain the expression for the Gain Bandwidth product of a transistor.
High Frequency Transistor Circuits
143
1. Giacoletto model
2.
The parameters are of different units (Hybrid)
in 1t - shape.
3.
Fictitious terminal
4.
gm
5.
1000
I
ole
oVB'E VCE
=:
=:
n ,U, constants etc. The shape of the circuit is
K.
6. ~'e =: gm / hr.
7.
hfe Vr
I Ie I
=:
h,e
8. becomes unity
9.
J;. =
hr.olp
10. C e =: gm/21t
J;..
11. 2.5 pf.
- 12. Incremental resistance.
13. rb••
=:
hre /
~
=:
200' O.
14. Picofarads
15.
J;.
=:
hfelp
16. Class A, B, AB, C
17.
r bb •
B
Ce
E -----'-----''-------'-----'---- E
18. Frequency at which short circuit current gain is unity.
Electronic Circuit Analysis
144
19. Tradeoffb/w gain and BW.
20. Capac!tances.
lId
21. 8", directly depends on current. or gm = V
T
22. 8", inversely proportional to temp.
23.
~'e =
gm => rb'e =
hCe •
hCe
gm
24. Cde a. current a. T"
W2
25. C -g - De
m
2DB
2
W
26. 21tf - - « l o r f« 3fT or f=-3
60 B
4
27. CC
28. input time constant.
29.
~=
hI
Ie
gm
2 (C
7t
e+
C )
c
IIeI
or gm = 26 mt.
UNIT - 4
Power Amplifiers
In this Unit,
• Power amplifiers - Class A, Class B, Class C, Class AB and other types
01 amplifiers are analyzed.
• Advantages and Disadvantages of different types are discussed.
• Thermal considerations and use of heat sinks is also explained.
146
4.1
Electronic Circuit Analysis
Introduction
When the output to be delivered is large, much greater than m W range and is of the order of few
watts or more watts, conventional transistor (BJT) amplifiers cannot be used. Such electronic
amplifier circuits, delivering significant output power to the load (in watts range) are termed as
Power Amplifiers. Since the input to this type of amplifier circuits is also large, they are termed as
Large Signal Amplifiers. In order to improve the circuit efficiency, which is the ratio of output
power delivered to the load Po to input power, the device is operated in varying conduction angles of
0
0
0
360 , 180 less than 180 etc. Based on the variation of conduction angle, the amplifier circuits are
classified as Class A, Class B, Class C, Class AB, Class D, and Class S.
4.1.1
Power Amplifier
Large input signals are used to obtain appreciable power output from amplifiers. But if the input signal
is large in magnitude, the operating point is driven over a considerable portion of the output characteristic
of the triinsistor (BJT). The transfer char~cteristic of a transistor which is a plot between the output
current Ie and input voltage VBE is not linear. The transfer characteristic indicates the change in ic when
Vb or IB is changed. For equal increments of VBE, increase in Ie will not be uniform since output
characteristics are not linear (for equal increments ofVBE , Ie will not increase by the same current). So
the transfer characteristic is not linear. Hence because ofthis, when the magnitude of the input signal is
very large, distortion is introduced in the output in large signal power amplifiers. To eliminate distortion
in the output, pushpull connection and negative feedback are employed.
I~IB
1
IB2
c
3
IBI
-VCE
Fig. 4.1 Output characteristics of BJT in CE mode
For simplicity let us assume that the dynamic characteristic of the transistor is linear.
Ie
r
Q
operating point
, I
Fig. 4.2 Transfer characteristics of B..iT
4.1.2
Class A Operation
If the Q point is placed near the centre a/the linear region a/the dynamic curve, class A operation
results. Because the transistor will conduct for the complete 360°, distortion is low for small signals
and conversion efficiency (11) is low.
Power Amplifiers
4.1.3
147
Class B Operation
o
ic
i
o
--Vb.
21t
Fig. 4.3 Transfer curve
FC!r class B operation the Q point is set near cutoff. So output power will be more and conversion
efficiency (ll) is more. Conduction is only for 180°, from 1t - 21t. Since the transistor Q point is beyond
cutoff, the output is zero or the transistor will not conduct. Output power is more because the complete
linear region is available for an operating signal excursion, resulting from one half of the input wave.
The other half of input wave gives no output, because it drives the transistor below cutoff.
4.1.4
Class C Operation
Here Q point is set well beyond cutoff and the device conducts for less than 1800. The conversion
efficiency (ll) can theoretically reach 100%. Distortion is very high. These are used in radio frequency
circuits where resonant circuit may be used to filter the output waveform. Class A and class B
amplifiers are used in the audio frequency range. Class B and class C are used in Radio Frequency
range where conversion efficiency is important.
--Vb.
Fig. 4.4 Transter curve
4.1.5
Large Signal Amplifiers
With respect to the input signal, the amplifier circuits are classified as
(i) Small signal amplifiers
(ii) Large signal amplifiers
4.1.6
Small SlgU.1! Amplifiers
Here the magnitude of the input signal is very small, slightly deviating from the operating point. But
always the operation is in the active region only. The characteristics of the device can be assumed to
be linear. We "ali draw· the equivalent circuit and analyse the performance. The magnitude of
the signal m:ly be few mY, in single digits. The operating point or Quiescent point Q swings
with the input signal. Because the input signal magnitude is small, the operating point is in the active
region only.
148
4.1-.7
Electronic Circuit Analysis
Large Signal Amplifiers
Here the magnitude of the ~nput signal is very large and deviation from the operating point on both
sides is very wide. So bec~~se of this, the device performance cannot be assumed to be linear.
Because of the large swing Of the input signal, the non linear portion of the transistor characteristics
are also to be considered. Hence the linear equivalent circuit analysis is not valid. So for large signal
amplifiers only graphical analysis is employed.
Power amplifiers, class A, class B, class C amplifiers, push-pull amplifier are of this type.
Large signal amplifiers are used where the output power requirement is large. Ifwe use small signal
amplifiers, the number of stages to be cascaded will be large, complicating the circuit.
Factors to be considered in large signal amplifiers:
1. Output power
2. Distortion
3. Operating region
4. Thermal considerations
5. Efficiency (11)
Amplifier circuits may be classified in terms of the portion of the cycle for which the active
device conducts.
It is one, in which the active device conducts for the full 360°. The device is
Class A
biased in that way.
0
Class B
Conduction for 180
0
Conduction for < 180
Class C
0
0
Class AB :
Conduction angle is between 180 and 360
4.2
Class A Power Amplifier
The circuit for class A amplifier considering only load resistance
RL
is as shown, in Fig. 4.5.
Fig. 4.5 Class A power amplifier
There are two types of operations :
1. Series fed
2. Transformer coupled
4.2.1 Series fed
There is no transformer in the circuit. RL is in series with Vcc. There is DC power drop across RL .
Therefore efficiency (11) = 25% (m~imum).
I
Power Amplifiers
4.2.2
149
Transformer Coupled
The load is coupled through a transformer. DC drop across the primary of the transformer is
negligible. There is no DC drop across RL . Therefore 11 = 50% maximum.
Vyand Iy are the root mean square (rms) values of voltage and current.
Ic = Iy
;'~ -.Iy~==--+--~_
o
rot
I
mm
2
1
1
I<..--V-I:------+-IV---....,.lv--~>OO""";~v c = Vy
rom ~ __ L~ __ ~ -=ax_. __ _
1
t
::::::'1
1
I
I
Fig. 4.6 Output characteristics
In class A amplifier, the conduction is for full 360°. Therefore the operating point lies in the
active region only. Let us assume that the static output characteristic of the transistor are ideal and
linear. So if the input is a sinusoidal signal, then the output will also be sinusoidal.
Let us use the subscripts y for output and x for input. Therefore ic
So subscript y is used. Input is on x-axis. So subscript 'x' is used.
The output power Py can be found graphically.
Vy = (rms) output voltage
Iy = (rms) output current
Subscript 'y' for output
Subscript 'x' for input
P = Vy Iy = 1/ RL'
1m
=
Inns
=
Peak value fo the current = (
1m /..[2
1m
Iy
=
..[2
=
Imax - Imin
2..[2
=
(Imax - Imm)
=
Peak to peak value
I _
I
p p
m
= I =-y
Iy
2.[i
Imax - Imin )
2
=~.
Output is on y-axis.
Electronic Circuit Analysis
150
_ Vm I m
- ( 1i·1i
4.2.3
J
Efficiency of Amplifier Circuits
Let Vyy is the DC voltage being supplied to the circuit and ~ is the DC current drawn by the circuit.
Therefore the DC power input to the circuit is Vyy. Iy. Let RL be the load resistance. Therefore DC
power absorbed by the load is
RL+ Iy Vy) where ~ and Vy (with small subscripts y) are the rms
current and voltages absorbed by the load and Iy (capital Y) is the DC current absorbed by the load.
In addition to the DC drop across the load and AC drop across the load there is thermal power
dissipation Po across the device, since it gets heated. According to the law of conservation of
energy, the input power should be equal to AC power, + DC power loss across the load and thermal
dissipation.
(1/ .
Vyy Iy
=
1/, RL + Iy Vy + Po
Vyy . Iy = Total input power
~2 . RL =
DC power drop in the load
Iy.V y = AC power in the load
But
Vyy = Vy + Iy RL
V Y = DC voltage, Iy = DC current
PD = (Vy + ~ RL) Iy = 1/ RL + Iy Vy + Po
PD = Thermal power dissipation.
Po = Vy Iy - Vy iy
If the load is not a pure resistance, Vy Iy should be replaced by Vy Iy Cos O. The total AC input
power + Input DC power = DC drop across the load + AC output voltage + thermal dissipated power.
Now if there is no AC output power i.e, the device is not conducting, then the rest of the power
should be dissipated as heat. Therefore if AC power output is zero, ie., AC input signal is zero, then
PD b maximum. and has its maximum value Vy Iy. Therefore the device is cooler when delivering
power to a load than when there is no such AC power transfer. When there is power drop across the
device itself, it gets heated.
4.3
Maximum Value of Efficiency of Class A: Amplifier
Certain assumptions are made in the derivation, which will simplify the estimation of the
efficiency (YJ). Because of this some errors will be there and the expression is approximate.
Power AmpUfiers
151
The assumption is that the static output characteristic of the transistors are equally spaced, in
the region of the load line for equal increments in the base current. If ib is increased by 1 !lA, ic will
increase by 1rnA, and if ib is increased by 3 !lA, ic will increase by 3mA. Thus for the load line shown
in the Fig. 4.7 the distance from 1 to Q is the same as that from Q to 2.
iy
I
max
i
1m
I
IY1+--I
I
o
V max
v
y
Fig. 4.7 TransferCurve
In the case of transformer coupled amplifier, supply voltage is only V y and not V max'
since the DC drop across transformer can be neglected. In the case of series fed amplifiers, supply
voltage V yy is V max'
~ = 1m
1m is the current corresponding to the operating point.
Vm is the voltage corresponding to the operating point.
Vm=lmZm=
Vmax -Vmin
2
The general expression for conversion efficiency is
Signal power delivered to load
- DC power supplied to output circuit
11 - ---="---"---------- x 100
Pac = Vm · Im/2
Poc
=
Vyy Iy
V m is the peak value or maximum value. Since rms value =
=50 Vm 1m %
Vyyly
Vm
.fi
Electronic Circuit Analysis
152
V
m
=
Peak value
1
2 (Peak to peak value)
= -
_ 50(Vmax - Vmin )Iy
11Vyy Iyx2
25 (Vmax - Vmin )
%
Vyy
-----=~
V I /2
11 =
m m
Vyy Iy
x 100%
1m = Iy ' since transistor will not conduct, if Imm = o.
P =
o
Ie
=
(VCC - VCE)I
C
-'--'=--.=!::"-'---''''-
2
Vee -VeE
R'
RL= Load resistance referred to primary.
L
Po = (Vee - VeE?/ 2
Poe = Vee x Ie
R{
~ Vee [Ve\-~VeE 1
P
_0_ x 100
P
DC
= 50 {I- VCE}
VCC
If it is a transformer coupled amplifier, V y is the DC voltage. Since Q point is chosen in the
middle of the load line, graphically,
Vy =
Vmax -Vmin
2
for a transformer coupled amplifier, there is no DC drop across the transformer.
Power Amplifiers
153
VCC-::::,VCE=Vy
DC input power = Vy' ~
Vyy
=
Vmax + Vmin
2
25tVmax - Vmin)x 2
T)
(Vmax + Vmin )
50tVmax-Vmi)
=
(Vmax + vmiJ
Vyy will be the quiescent voltage itself for transformer coupled amplifier. Since there is no DC
voltage drop across the transformer.
If Vmin = 0, maximum efficiency = 50% for class A transformer coupled amplifier.
For series fed amplifier, Vyy = Vmax = 2 V m
v
- 25 t max - Vmin ) OJ: If V
- 0
.
- 250J:
V
/'0.
max ,maXImum T) /'0
max
Therefore for a transformer coupled amplifier conversion" is twice. (50% compared to 25%
for series fed amplifier)
T) -
4.4
Transformer Coupled Amplifier
In AC amplifier circuits, the input AC signal should be coupled to the amplifier and output of the
amplifier should be coupled to the load resistance. The coupling device should be such that, it allows
only the AC signal to the amplifier circuits and blocks the DC components present in the signal
generator, because we are interested in amplifying only AC signals. For this purpose a capacitor can
be used for coupling.
Fig. 4.8 Coupling in amplifier circuits
Types of coupling "
1. Capacitor coupled amplifier
4. Direct coupled amplifier
2. Transformer coupled amplifier 3. RC coupled amplifier
5. Inductor or Tuned amplifier
Es is the AC signal generator and Rs its source resistance. C I and C2 are the coupling capacitors.
They are chosen such that, for the lowest frequency signal to be amplified, Xc and Xc are
I
2
Electronic Circuit Analysis
154
short circuits. But because of these reactive coupling elements, as signal frequency decreases Xc
increases. Hence there will be large voltage drop across the capacitor and so the actual input to the
amplifier reduces and hence gain decreases. Similarly at high frequencies because of the shunting
capacitance C s' gain falls. Therefore there is a particular frequency range in which gain is of desirable
value only.
Instead of capacitors, transformer can also be used for coupling.
/I
/I
Fig. 4.9 Transformer coupling
Transformer does not respond to DC. Therefore only AC signals fro¢. source to the amplifier
circuit and from the amplifier to the load will be coupled. But what is the advaptage of the transformer
coupling? Suppose, the load resistance RL is very small : : : 40, 80 or ISO as in the case of a loud
speaker. The output impedance Ro of the transistor amplifier is much larger (for common emitter and
common base configuration A v ' Ai »1). Therefore impedance matching will not be there and so
maximum power will not be transferred. Even if capacitive coupling is used, impedance matching
cannot be achieved. But this can be done using a transformer.
[ :~ ) = Turns ratio
of transfonner
N 1 = Number of turns on primary side.
N2
=
Number of turns on secondary side.
R, is much larger than RL" Therefore
[~~) > I or the transfonner that should be used should
be a Stepdown transformer. Therefore the output voltage at the secondary of the transformer will be
much smaller compared to the input voltage since stepping down action is taking place. [This is the .
case with class A and class B power amplifiers in the case of lab experiment]. But the current
amplification will be there and because of Z matching, maximum power will be transferred to the
load. Similar to resistive capacitor coupled amplifier, we have the frequency response which depends
upon the inductance of the primary and secondary. The transformer on the primary side is chosen
such that the sotirce resistance of the generator matches with the input Z of the amplifier circuit.
Transformer coupled amplifiers are used in low audio frequency range only because at higher
frequencies the XL of the transformer will be large and so the gain falls.
Another advantage with the transformer coupled amplifiers is the AC current passing through
the load resistance RL results in only wastage of power, since we are interested in only AC output
Power Amplifiers
155
power. Moreover passing DC current through the loudspeaker coil is not desirable since it produces
hum or noise. Therefore if transformer coupling is done, DC component of current passing through
the transformer can be avoided.
II
Es
Fig. 4.10 Transformer coupled amplifier
(For a transformer with usual notation.)
R 1, R2, RE ,\re chosen depending upon the biasing point. CE is emitter bypass resistor. Transformer
T 1 is chosen to match ~ of the circuit with Rs and transformer T2 is chosen for Ro of the circuit to
match with RL •
C2 is also a bypass capacitor. For AC it is short circuit.
The equivalent circuit, in terms of h-parameters neglecting the biasing resistors and capacitors,
also neglecting the input transformer and considering base and emitter as the input ports.
hIe
E~----~------------~--~--------~~----~
Fig. 4.11 Equivalent Circuit
Electronic Circuit Analysis
156
The transistor is replaced by its h-parameter equivalent circuit. The load resistance RL is
referred to primary and so R~
RL
=2
where n
N2
= N. Since it is a stepdown transformer, n < 1; Lp
n
1
is the inductance of the primary winding (Since we are considering load referred to primary).
4.4.1
Mid Frequency Range .
In the mid frequency range, the inductive reactance X LP is high. fis large and so it can be regarded
as an open circuit (or very large compared to R~). Therefore it will not affect the response. hfe Ib is
the current source. When impedance matching is done, the output Z of the circuit and the load
resistance RL will be equal. Therefore the current will get divided between RL and the circuit equally.
The total current is hfe lb. Therefore the current through the primary of the transformer (or in other
words the current through the collector circuit) is
Thfi .Ib
=
Icl ·
But since T2 is a srep down transfonner the current through RL will be stepped up by (
due to transformer action
~
N2
> 1.
.[=~l
:. Current gain
4.4.2
Voltage Gain
But
Vc
NI
VL
N2
-=-
VC = voltage on primary side of transformer
VL = voltage on secondary side of transformer
V
hfe RL
Nt
A v = -L= - - x - x Vb
2
hie
N2
=~ )
Power Amplifiers
157
In the low frequency range shunting effect of Lp will reduce the effective load resistance.
Lower 3db frequency is reached when,
2 1t fl Lp = R
R
or
II = 21tL
P
where R is the parallel combination of
foe
and
[~l2. RL
N2
VI
Fig. 4.12 Class A power amplifier circuit
The transformer coupled amplifier circuit is similar to a circuit, like this. Instead of having
coupling capacitors C I and C 2, we have transformer, coupling. The primary of transformer T2 acts as
R.c. C2 across R2 helps in making Emitter of transistor at the ground point for AC. Because of CE,
emitter is at ground potential for AC. Therefore secondary voltage of transformer is applied between
base and emitter of transistor.
4.5
Transformer Coupled Audio Amplifier
Audio amplifier
Video
R.F
40Hz to 20 KHz
5-SMHz
20KHz
Classification of Radio Waves.
Very low frequency(VLF)
Low frequency (LF)
Medium frequency (MF)
\
High frequens;y (HF)
VHF
Ultra high frequency (UHF)
Super high frequency (SHF)
10-30 K Hz
30-300 K Hz
300-3,000 K Hz
3-30 MHz
30-300 MHz
300-3000 MHz
3000-30,000 MHz
An amplifying system usually consists of several stages in cascade. The input and intermediate
stages operate in a small signal c1ass-A mode. Their function is to amplify the small excitation
to a large value to drive the final device. This output stage feeds a transducer such as CRT, loud
Electronic Circuit Analysis
158
speaker, servo motor etc. So the output stage must be capable of delivering a large voltage or current
or large power. Bias stabilization techniques and thermal runaways are very important with power
amplifiers.
If the load resistance is connected directly in the output circuit as shown in FigA.13 (a), the
quiescent current passes through RL • This results in waste of power since it won't contribute to the
AC power signal. In the case of loud speakers it is not desirable to pass DC current through the voice
coil. So an arrangement is to be made using an output transformer.
II
(a)
4.5.1
Ce
(b)
Fig. 4.13 Transformer coupled amplifiers
Impedance Matching
To n.:ansfer significant power to a load such as loud speaker with a voice-coil resistance of 5-150 it
is necessary to use an output matching transformer. The impedance matching properties of an ideal
transformer are :
VI
Nl
V2
N2
-=-;
Let
N2
NI
=
n turns ratio.
If N2 < N I the transformer reduces the output voltage, and steps up the current by the
same ratio.
Impedance matching is required because the internal impedance of the device will be
much higher than 5-150 of voice coil of a speaker. So power will be lost. Hence output matching
transformer is required.
Power Amplifiers
159
fv1
f
=
Effective input resistance R{
=
Effective output resistance
V
4.5.2
R~
2
Maximum Power Output
j
Is =-25 rnA
-2.0
Ic (A) -1.5
Collector Current
-0.5
o
o
-5
--...;')~
-10
-15
Collector voltage VCE
Fig. 4.14 (a) Output characteristics
V
lbe
I
40rn
I
(1\
.,---
\ ) wt
Q
I
VSEj
T
o
-5V
-15 -20
-20
Base current ~ is rnA
Fig. 4.14 (b) Input characteristics
To find n, for a given RL , so that power output is maximum is solved graphically. First
. pomt
. Q'IS Iocated I c = Vee
operatmg
Rc ;
Pc = collector dissipation
Power
output
t
Load Resistance ~'.
Fig. 4.15 Variation of Po with R; L
Electronic Circuit Analysis
160
Ve = quiescent collector voltage. Peak to peak voltage mu~t be limited to a suitable value such
that there is no distortion. From the input characteristic IBmax to IBmin can be noted. A series of load
lines are drawn through 'Q' point for different values of R~. From these two graphs, power output
versus load resistance RL is drawn from the graph, R~ is chosen that power is maximum and distortion
is minimum.
4.5.3
Efficiency
Suppose the amplifier is supplying power to pure resistive load.
Power input from DC supply
=
Vee Ie
Ie 2 R] + ie 2 Ve
R] is static load ie and Ve are rms output current and voltage. If Po is average power dissipated
by the active device.
Power absorbed by output circuit
=
+
(a)
(b)
Fig. 4.16 Power amplifier circuit
Vee Ie
But
Vee
=
=
Ie2 R] + ie ve + Po
Ve + Ie R];
Po
=
Vee· Ie - Ie 2 R] - Ie 2 Ve
Po
=
Ve Ie + R] Ie 2 - ie ve
Po = Ve Ie - ie ve
If the load is not pure resistance, ie ve must be replaced by ie ve Cos
power factor of load.
4.5.4
e, where Cos e is
Conversion Efficiency, TJ
An amplifier is essentially a frequency converter, changing DC power to AC power.
A measure of the ability of an active device to convert DC power of the supply into AC power
delivered to the load is called conversion (l]) or theoretical efficiency (l]). It is also called collector
circuit (l]) for transistor amplifier.
Signal power delivered to load
T)=----'''-----''--------- x 100%
DC power supplied to input circuit
In general,
Power Amplifiers
161
I max
t
Ie
I
m~m~~~~~~~~__~
vc
V mm
v
1m'.
v
Fig. 4.17 Transfercurve
where Bo and B\ are constants in the expression
ic
=
Ic + Bo + B\ Cos eot + B2 Cos 2eot + .....
Expression for instantaneous total current.
If distortion components are negligible,
Tl=
lv
2 m
V
Ce
m
I
x
100%
C
Maximum Value of (Tl)
In the case of series fed amplifiers, the supply voltage Vcc is equal to Vmax' In the transfonner
coupled amplifier Vcc is equal to the quiescent voltage Vc'
Under ideal conditions,
Ic= 1m
V
and
Vm =
Tl
=
max
-V.
mm
2
Vm.I m
50 V. I
cc c
max
50X(V
~ Vmin }Ic
Vccl c
For series fed amplifier, Vmax = Vcc
:. Tl for series fed amplifiers
162
Electronic Circuit Analysis
= 25~Vmax - Vmin) %
V
max
Maximum possible value = 25%
In the case of transformer coupled amplifier.
V =V =
ce
11
c
=
50 (
V
+V.
max
mm
2
Vmax - Vmin)
Vmax + Vmin
%
So the Maximum Possible Value of 17 is 50% for transformer coupled amplifier
Thus transformer coupled amplifier have twice the maximum 17 compared to series fed amplifiers.
For transformer circuits occurs near saturation, therefore Vmin « V max and 11 can be 50%.
PNP Transistor Amplifier with AC Signal
Emitter is forward biased. Collector is reverse biased.
AC is superimposed at the input, we get AC output across Rv (Fig. 4.18)
output
AC
voltage
Fig. 4.18 Circuit with PNP transistor
4.6
Push Pull Amplifiers
RI and R2 are provided to prevent cross over distortion. Because ofRI and R2 the B-E junctions of the
two transistors are forward biased so that cut in voltage Vr will not come into the picture. But because
of RI and R2, the operation will be slightly class AB operation and not pure class B operation. For a
given transistor, the dynamic characteristics are not exactly linear, that is, for some changes in the
values of ib, ic will not change by the same amount that is. If ib is increasing by 5 ~, ic increases
by lmA. For 10 ~ increase in ib, ic will not increase by 2 rnA but something different. So the graph of
Fig. 4.19 (a) Push Pull amplifier circuit
Power Amplifiers
163
i., Vs ie is nonlinear. Therefore for uniform changes in the input, the output will not change
uniformly. Hence,distortion will be introduced in the output waveform. This can be eliminated by
pushpull connection.
The efficiency (11) for class A amplifier is 25% and for transformer coupled class A amplifier is
50%. Therefore class A amplifier because of poor 11 is used for low output power requirement and
where conductance should be for complete 3600 • (eg. for the driver stage of the last power stage).
Suppose for the transistor QI' the input (base current) is a Cosine waVe XI = Xm cos rot. The
output current at the collector i l = Ie + Bo + BI Cos rot + B2 Cos 2rot + B3 Cos 3rot where Ie is the DC
current due to biasing, Bo is the DC component in the Fourier series ofthe AC input, BI is fundamental
component Cm (Transformer TI provides phase shift to the inputs. T2 joins the two outputs. For the
second transistor, the input is given from the centre tapped transformer which introduces a phase
shift of 1800 •
x2=-xl =Xm{Cos(rot+1t)}
The output current of this transistor i2 is obtained by replacing rot by (rot + 1t) in the
expression for i l .
i.e.,
i2 (rot) = i l (rot + 1t)
i2 = Ic + Bo + BI Cos (rot + 1t) + B2 Cos 2 (rot + 1t) .... .
= Ic + Bo - BI Cos rot + B2 Cos 2rot - B3 Cos 3 rot .... .
Therefore i I and i2 are out of phase by 1800 • So they flow in the opposite direction through the
output transformer primary windings. Therefore the total output current i is proportional to (il - i2).
Since the net output current depends on the turns ratio of the transformer i is the current flowing
through Rv
i = K (il - i2) = 2 K (B I Cos rot + B3 Cos 3 rot) .....
This expression shows that all the even harmonic terms B2 Cos 2 rot, B4 Cos 4 rot are eliminated.
The only harmonic component predominant is B3 Cos 3 rot, the III (third) harmonic terms. Higher
harmonics can be neglected. BI Cos rot is the original signal. Therefore Harmonic distortion will be
less for pushpull amplifiers. This is under the assumption that both the transistors have identical
characteristics. If not, some even harmonics may also be present.
Pushpull amplifier is said to possess mirror symmetry.
Mirror symmetry means, mathematically,
i (rot) = - i (rot + 1t)
The output current i for pushpull amplifier is,
i = 2 K (B I Cos rot + B3 Cos 3 rot + ..... )
If rot is replaced by (rot + 1t), the above equation holds good.
This is also called as halfwave symmetry. It means that the bottom loop of the wave when
shifted by 1800 along the axis will be the mirror image of the top. This is so because only odd
harmonic terms are there in the output.
The maximum instantaneous reverse voltage across each transistor occurs when it is not
conducting and is equal to 2 VI. Because when Q I is conducting, maximum VCE = 0 and so voltage
Electronic Circuit Analysis
164
across the upper half winding of output transformer primary is Vcc. Due to induction some voltage
will appear across the lower half also. Q2 is not conducting. Therefore the transistor voltage across
Q2 collector and emitter is Vcc + V Ce = 2 V CEo
It is called as push pull amplifier since, the input to the two transistors are out of phase by
180°. (Since centre tapped transformer is used). Therefore when one transistor is conducting the
other is not or when the output current of one transistor is increasing, for the other it is decreasing.
This is known as push pull action. When one transformer current is being pushed up, the other is
being pulled down.
4.6.1
Class B Amplifiers
A transistor circuit is in class B operation, if the emitter is shorted to base (for DC). The transistor
will be at cut off. Therefore in the circuit for class B push pull amplifier, R2 should be zero. The
conduction angle is 180°.
4.6.2
Advantages of Class B Push Pull Circuit AMPLIFIER
I. More output power; YJ = 78.5%. Max.
2. YJ is higher. Since the transistor conducts only for 180°, when it is not conducting, it will
not draw DC current.
3. Negligible power loss at no signal.
4.6.3
Disadvantages of Class B Push Pull Circuit AMPLIFIER
1. Supply voltage Vcc should have good regulation. Since if Vcc changes, the operating point
changes (Since Ic changes). Therefore transistor may not be at cut off.
2. Harmonic distortion is higher. (This can be minimized by pushpull connection).
Therefore Class B amplifiers are used in a system where the power supply is limited, and is to
be conserved such as circuits operating from Solar cells or battery, Battery Cells, air borne, space
and telemetry applications.
4.6.4
Conversion l'J
P
=
1m x Vm
° .fi.fi
= 1m
Vm
2
=
1m (V
_V . )
2
CC
mm
Vm = Vcc - V mm (Since operating point is chosen to be at cut off Vcc = Vmax)
Because in pushpull circuit, there are two transistors conducting, each for 180°.
Therefore total conduction is for 3600. (:. Po
= 1m~m ) Corresponding to Q point,
the voltage is VCEo Neglecting the dissipation across emitter, Vcc ::: V CEo
P OC = loc . V cc. DC current is drawn by the transistor only when it is conducting.
Vcc is always present. One transistor conducts for 0-1t only. loc drawn by each transistor is the
average value of half wave rectified DC (equal to Im/1t). But there are two transistors.
:. Total
loc
=
21m
-1t-
Power Amplifiers
165
- V
21m
POCcc·-1t
If
4.6.5
Dissipation of Transistors in Class B Operation
The DC input power to the transistors in class B configuration is
Poc =
2lm·Vee
1t
[Since the transistor is conducting for 180° only. So it draws DC current only during that
I
period. Therefore average value of ic is --.!.!!... There are two transistors each conducting for 180°,
1t
from
0 - 1t
and 1t -
21t
respectively).
1t
~rot
Fig. 4.19 (b) Current cycle
Total DC current = 2 Im/1t
21m
Poe
But Im
=
Vm
I
=
--.Vee
1t
where R{ is the effective load resistance of the Circuit, without considering the
RL
secondary of the transformer.
Electronic Circuit Analysis
166
pr!~~
G S''::~
Fig. 4.20 Transformer on load side
The collector dissipation Pc (in both transistors) is the difference between the power input
(PI or Poe) to the collector circuit, and the power delivered to the load. (Both are in watts). Though
DC and AC powers.
Output power delivered to load Po'
Po~ ~2 ~(Ar/R~
Vm 2
,
2RL
Pc = Pi - Po =
[~. vcc,Vm]
-( v~ J
RL
2RL
1t
(Vm is the peak value ofthe AC input).
The above equations shows that at no AC signal, (Le., Vm = 0) the collector dissipation is zero,
and as the signal magnitude increases, Pc increases. As Vm increases Pc also increases, Pc is maximum
when Vm =
2VCC
--1t-
Vee /4
~
Fig. 4.21 Power output
A graph can be plotted between Pc and Vm
The maximum dissipation
Pc max =
Pc is maximum, when Vm =
2VCC
- - .1t
Po is maximum, when Vm = Vcc
Power Amplifiers
167
2
Po (max) =
VCC
--,-
2RL
V2
Since P=~
o 2R
L
Po is maximum
4
Pc
Ipe
(max) =
2
(Po max)
1t
(max)
=
0.4 Po (max)
I
If we want to deliver 10-W of output by a class B pushpull amplifier, the collector of the
transistor or the collector dissipation should be 0.4 x 10 = 4W. This is for the entire circuit. Therefore
each transistor (Since there are two transistors in class B pushpull) should be capable of dissipating
2W of power as heat.
4.6.6
Graphical Construction for Class B Amplifier
JICI
for class B, Q point is chosen as shown
---+ V
eel
Fig. 4.22 Class B operation
4.6.7
Distortion
Let ib ' Vc, Vb be the input characteristic of the first transistor and ib ' V s' Vb is the input characteristic
of th6 second \ransistor. Vy is the cut)n voltage. These are t6e two tiansistors of the class B
pushpull amplifier. Now the base input voltage being given to the transistor is sinusoidal, i.e., base
drive is sinusoidal. So because of the cut in voltage, eventhough input voltage is present, output will
not be transmitted or there is distortion in the output current of the transistor. This is known as
crossover distortion. But this will not occur if the base current drive is sinusoidal. Since in the
graphical analysis the input current is taken in the I quadrant. No distortion if the operating point is
in the active region. Cross-over distortion can also be eliminated in class AB operation. A small stand
by current flows at zero excitation. The input signal is shifted by constant DC bias so that the input
signal is shifted by an amount V1"
Electronic Circuit Analysis
168
v~
I
---t----
Q2
1
I
I
ib
2
rot
Fig. 4.23 Cross Over Distortion
Table 4.1 Comparison of amplifiers based on the type of Coupling.
Direct coupled
Transformer coupled
Frequency range :
D.C to medium range High
f\ (Lower cutoff
frequency
o Hz (D.C)
50-100 Hz and above 100 Hz and above
f2 (Upper cutoff
frequency)
Limited
Can be more
Limited to A.F range
Cost
Less
No Rand C
No transformer
Medium
(Due to R and C)
High
(Due to transformer)
Size
Less
Medium
High
Frequency response
IZI
4.7
R.c. coupled
matching
Afl~
Not good
A.F. range
llc\
~f
Not good
W-
(due to t oftarisformer
spike occurs)
Excellent
Complimentary Symmetry Circuits (Transformer Less Class B Power Amplifier)
The standard class B push-pull amplifier requires a centre tapped transformer, since only one transistor
conducts for 180°, so that if two transistors were to conduct for complete 360°, there should be a
centre tapped transformer. Otherwise there should be a phase inverter. Complementary symmetry
circuits need only one phase. They don't require a centre tapped transformer. But their requirement is
Power Amplifiers
169
a pair of closely matched. Oppositely doped (pnp and npn) transistors. Till recently it was different
to get such transistors. But now the technology has improved and pnp and npn transistors with
identical circuits, can be manufactured.
V
,
rv
Vee
1
Vcez
Fig. 4.24 Complimentary Symmetry
The circuit shows a basic complimentary circuits in class B. It is class B operation since the
operating point is at cutoff. Emitter and base are shorted or VBE = o. The input is capacitance
coupled. The output is direct coupled since output is taken directly across RL. One end of RL is
grounded with no input signal present. Both transistors won't conduct. Therefore current through
RL = o. When the signal (input) is positive going, the transistor Q, is cutoff (since it is pnp), base is
n type. Therefore. Emitter-Base junction is reverse biased). Q2 conducts, since it is NPN transistor,
base input is positive. So it conducts.
The resulting current flows through RL and develops a negative going voltage at point relative
to ground. When the signal is negative going Q2 goes off and Q, turns on. Current flows through RL
in such a direction as to make point positive with respect to ground. There is no DC current through
Rr- Hence an electromagnetic load such a loud speaker can be connected directly without introducing
saturation problems.
The difficulty with the above circuit is, the transistor will not conduct till the input signal
magnitude exceeds the cut in voltage Vv. So cross over distortion will be present output of the
Fig. 4.25 Output with crossover Distortion
So DC bias should be provided to overcome the threshold voltage for each base-emitter
junction. Therefore the circuit is as shown in Fig. 4.26(a). The voltage developed across R2 forward
biases both the transistors, E-B junctions. R2 is normally so small as not to produce any significant
loss in drive to Q2.
This circuit needs only one Vcc Q, is NPN transistor. Therefore its collector is reverse biased,
since VCE is positive.
Electronic Circuit Analysis
170
Q2 is PNP, its collector is negative with respect to VCE since grounded. Therefore its collector
is also reverse biased. The drop across R3 reverse biases the common base junction of Q2 and the
drop across RI reverse biases the Common Base junction of QI' This circuit requires only one DC
supply and is commonly referred to as the "Totem pole" configuration. When the input is positive, Q I
is turned on and Q2 is turned off. When the input goes negative, Q I turns off while Q2 conducts.
4.8
Phase Inverters
These circuits are used to drive push pull amplifiers since a pushpull amplifier requires two equal
inputs with 180° phase difference. The· centre tapped transformers are bulky and costly. Therefore
Phase inverter circuits with transistors are used.
Phase inverter circuits are also known as Paraphase amplifiers. The criterion is, from a single
input, we must get two equal outputs with a phase shift of 180°.
VOI =-V02
The circuit is as shown in Fig. 4.26(b), emitter followers with a collector load RI is used.
R3 and R4 are bias resistors. RI is the collector load. R2 is the emitter load. Output is taken
after the capacitor 'C' to block DC. Vi is AC input. The outputs at points 2 and 1 will be out of phase
Input +
v o---j I--+---H
+
I
t-----jH
(b)
(a)
Fig.4.26 Phase Inverter Circuits
by 180°. With RI = R2 output voltages will be the same. The output impedance of the circuit from
point 1 is that of a common collector configuration (Since output is taken across R2• So it is common
collector configuration).
The output impedance at terminal 2 is that of common emitter configuration. So both will not
be the same since RoE =1= Roc, since output voltages at points 1 and 2· will not be the same. So for
that, another transistor is used to match the gains and source impedances for common emitter
collector. There is phase shift of 180°. Therefore V02 will be with phase shift for common capacitor
configuration there is no phase shift. Therefore VOl is in phase. Therefore VOl and V02 are out of
phase by 180°.
Example: 4.1
The amplifier shown is made up of an NPN and PNP transistors. The h-parameters of the two
transistors are identical and are given as ~e = 1 Kn, lIre = 100, hoe = 0 hre = O.
Find overall voltage gainAv = VONi
Power Amplifiers
171
Both the transistors are in common emitter configuration. For Q2 the output is taken across
5 K!l the collector resistor Rc which is actually the load resistor.
2
lL
A~I
Rc I
=RLI =2kn
=lkn
r---------~~+v~
fig. 4.27 Circuit for Ex : 4.1
R
A
v2
=~= -5Kn =5
RC2
IKn
2Kn
RLI
A =--=-=2
vI
IKn
RC
1
Av =A vI xAv =5x2=10
2
Electronic Circuit Analysis
172
Example: 4.2
Design a class B power amplifier to deliver 30W to a load resistor RL
coupling. Vm = 30V
=
=
40 using a transformer
Vcc' Assume reasonable data wherever necessary.
Solution:
The power to be delivered is 30W. Assume 10% losses in the transformer windings, and design the
circuit for 20% over load i.e., even by mistake, if excess current is being drawn or even voltage
applied, the transistor must with stand this .
.. Po is taken as 40W. 30 + 7W (overload) + 3W (transformer losses)
.. The collector dissipation of the transistor
Pc (max) = 0.4 Po (max)
.. The transistor to be chosen must be capable of dissipating Pc (max) = 0.4 x 40 = 16 W
Pri~ ~ [I~ _n~
Fig. 4.28 Circuit for Ex : 4.2
V2
m
2RL
Po =
RL
(30f
=
2x40
RL = 11.250
RL is the resistance of transformer secondary referred to primary.
RL =
RL
[!i)2
N2
Vp Ip = Vs Is
Vp
IS
N 1'
-=-=Vs
Ip N2
.. The turns ratio of the output transformer is,
(Rl]
Nl = n =
N2
RL
V2
2
= (11.25)1/ = 1.7
4
Power Amplifiers
173
Peak collector current swing
1m
=
Vm = Vee = ~
RL
RL
11.25
=
2.666 Amperes
Example : 4.3
Oesign a class A transformer coupled amplifier, using the transistor, to deliver 75 mW of audio power
into a 40 load. At the operating point, IB = 250 IlA, Vcc = 16V. The collector dissipation should not
exceed 250 mW. RL' = 900 O. Make reasonable approximations wherever necessary.
VCE =
V
;C for biasing in normal amplifier. If it is transformer coupled, VCE :: VCe
Collection dissipation = Ie. VCE (Ic is the collector current at operating point).
or
VCE · Ic -= PDmax
VCC . Ic :: PD max
Because, the DC. drop across the transformer Vcc :: VCE can be neglected. The drop across
RE is small :: IV. Therefore VCE :: Vcc - V = 15 V.
Fig. 4.29 Circuit for Ex : 4.3
VCE = 16V. There will be some voltage drop across RE and the primary winding of the
transformer. Therefore VCE can be approximately taken as 15V.
Ie
Po min
=
VCE
250
=-
15
16.66 rnA
Assuming that transformer primary resistance negligible,
VE = Vee-VeE
= 16 - 15 = IV (DC)
IE :: Ie = 16.66 rnA (DC)
=
RB =
VE _ IV(OC)
IE - 16.66rnA
=
600
Electronic Circuit Analysis
174
At f= 50 Hz,
eE =
21tfX
=
E
1
- 53 ~f
2x3.14xSOx6-
R'=900Q
L
.. Transformer turns ratio
n
=
=~ = t~O) = 15
It is Germanium transistor, V BE = 0.2SV
VB=VE+VBE
= I + 0.25 = 1.2SV
Assuming that the current through R1 is 10 IB,
IR = 250 ~ x 10
1 =2.5 rnA
Neglecting the loading effect due to base of the transistor and assuming that IB, flows
through RL also
R _ VB = 1.25
2 - IRI
2.SrnA
= 0.5 KQ
R1
=
Vee - VB
IRI
4.9
=
16-1.25
=
5.9 KQ
2.5 rnA
Class D: Operation
These are used in transmitters because their efficiency (11) is high::: 100%.
o
1+
VOp
·A ,
J_ r~~--rot
Vo
Fig. 4:30 Class D amplifier circuit
175
Power Amplifiers
A pushpull connection of two transistors in common emitter configuration of complementary
transistors (one pnp the other npn) is employed. When the input is positive, Tl is cut off and T2
saturates. During the negative half cycle of the input, T I saturates and T2 is cutoff. Therefore the
output voltage is a square wave with voltage changing between 0 and Vcc'
The dot convention for transformer is, when input is positive, the dotted end of the primary is
positive. At the same time, the dotted end of upper secondary winding is positive and dotted end of
lower secondary winding is positive. So when the input is positive, Tl base which is n type (pnp) gets
positive voltage. So T I is cutoff. Therefore V = Vcc. When input is negative, T 2 base which is p
type (npn) will get negative voltage. So T2 is cutoff T I saturates.
In this circuit, each transistor is saturated for almost 1800 of the cycle. So each transistor acts
like a switch rather than like a current source. When the transistor saturates, the power dissipation.
PD = VCE (sat) Ic (sat)
It is very small, since VCE (sat) is near zero. When the transistor is cutoff, PD :::: O. Therefore
average power dissipation over the cycle is very small. Therefore TJ :::: 100%. ~cc
The output of the collectors of transistors, is a square wave with 0 - V cc voltages. This is
given to a series resonant circuit. So the output will be a sine wave (like oscillator circuits).
4.10
Class S: Operation
Switching regulators are based on class'S' operation.
In class S operation, a string of pulses are used as the input signal. The pulses have a width
'W', and a period 'T'. Therefore duty cycle
W
=
T
=
D.
4.10.1 Circuit
L
+~U-D+
c
oiWH-T~
Fig. 4.31 Class S amplifier
The transistor is an emitter follower driven by a train of pulses. Because of the V BE drop, the voltage
driving the LC filter is a train of pulses with an amplitude of
Vcc - V BE ·
If XL > Xc
V DC = D (Vcc - VBJ
where
W
D = - = Duty cycle
T
Electronic Circuit Analysis
176
The higher, the duty cycle, the ,larger, the DC output. By varying the duty cycle, we can
control the a.c. output. So this is class'S' operation. Because the transistor is cutoff or in saturation
its power dissipation is much lower than that in a series regulator. So heat sinks can be small.
Diode rectifies and L, C combination filters the output. So the output is rectified
and filtered.
Class A: Conduction of plate current is for complete 3600 , it depends upon operating point.
Class B : Conduction of plate for only 1800 because the grid is more negative during negative
cycle of the signal.
Class C :Conduction is for less than 1800 .
Class AB : Conduction is between 360 and 1800
e~
c
Fig. 4.32 Input output waveforms
Class A
Less power
Lesser l'J
Less Harmonic distortion
Class B
More power
More l'J upto 78.5%
Harmonic distortion is more
Example: 4.4
Design a class A power amplifier to deliver 5V rms to a load of 8 Ohms using a transformer coupling.
Assume that a supply of 12V is available. The resistance of the primary winding of the transformer
also should be considered.
Solution:
1. First select a suitable transistor
The power output required
j"
(Vorms )2
RL
5x5
25
= -8- = 8 = 3.125W
Assuming a transformer efficiency, l'J, of 90%, we have, the power required of the amplifier
= Pol l'J
= 3.125/0.9 = 3.47W
Power Amplifiers
177
12V
0.8: 1
110
80
Fig. 4.33 Class A Power Amplifier
Therefore, we shall have to design the amplifier for 3.47W. Since the maximum efficiency of
the transformer-coupled power amplifier is 50%, the power dissipation capability of the transistor
should be at least 3 to 4 times the power required to be developed.
For the transistor, therefore, the.P d(max) should be
= 3.47W x 3
=about 10.41W
Let us sele'it a transistor, EC3054, for the purpose.
This transistor has
Pd(max) = 30W at 25°C
Ic(max)
=
4A
VCE(sat) = 1V.
2. Choosing Q-point
For transformer coupled amplifier, ideally, VCEQ = Vceo We shall assume the voltage across
the resistance RE as about 20% of the supply voltage, i.e.,
V E = 0.2
x
12 = 2.4V
Since VCE (sat) = 1V, and also to avoid the distortion near the saturation region, we shall take the
quiescent point voltage.
= VCEQ = about 2/3rd Vco giving us
VCEQ = 8V
The maximum swing available will be about 1V less (V CE (sat) = 1V) than the supply voltage of
12 volts.
Electronic Circuit Analysis
178
Hence for a power of 3.47W, we have,
Vp
Ip
..fi x ..fi
3.47 =
giving Ip = 1.21 Amps.
Therefore, the Q-point is at 8V, 1.21A.
3. Choosing RE
We have assumed voltage across the resistance RE as equal to 2.4 V, being about 20% of the
supply voltage VCC.
Therefore,
R
E
= VRE = 2.4V
ICQ
1.12A
=2.20.
which is the nearest available standard value of the resistance. Let us recalculate the voltage
across the resistance R E •
The voltage
VRE = 1.21A x 2.20 = 2.662 Volts
The power dissipation of the resistance
RE = (1.21 A)l x 2.20
= 3.22 Watts.
Hence, we select the resistance
RE = 2.2 Ohms, 10 Watts.
4. Thrns Ratio of Transformer
Secondary voltage = 6V (rms).
Let us calculate the primary voltage and, hence, the turns ratio. At Q-point, the DC voltage
across the primary is
V cc - V CEQ - (lCQ x RE)
= 12 - 8 - (1.21 x 2.2)
=
=1.4Y.
Giving DC resistance of the transformer
~nmary
= 1.4V/1.21A
= 1.16 ohms.
The equivalent resistance on the primary of the transformer is equal to
Vp
= -I-
p
= (11
- ~rimary
I 1.21) - 1.16
= 7.93 Ohms
Rae - ~rimary
179
Power Amplifiers
6
The turns ratio
7.93
= 0.76 ::: 0.8.
5. Choosing Resistance R. and
Assuming RB
=
~
10 times RE, for good stability, we have,
Rl xR2
R 1+ R 2
=:
..... (1)
10 RE
Also, since VE = 2.4V, VB = 3V
=
we have
3
or
R}
R2
Rl +R2
x 12
= = 3 R2
..... (ii)
From equations (i) and (ii) above, we have
n.
28.6 n.
R} = 88
R2 =
We select the nearest available values, as
n.
33 n.
R} = 100
R2
=
The power rating of these resistances are as under,
For
R} =
(VRl f
Rl
For
R2 =
Hence, we select, R}
=
(V B2
R2
'f
=
(18-3f
=
100
3
= 2.25W
2
33
=
0.27 W
100 Ohms, 5 Watts.
R2 = 33 Ohms, 1 Watts.
Let us calculate the maximum undistorted power available, which is equal to (Vp /
=
(1I1fi) x (1.21/ fi)
=
6.66 Watts, which is more than the required values.
fi) x (I P / fi)
Electronic Circuit Analysis
180
The circuit efficiency :
Useful power output
=
6.66 Watts.
=
14.52 + 1.082
Power input
=
15.60 Watts.
The circuit efficiency, is, therefore,
=
(6.66 / 15.6)
=
42.7%
x
100
Likewise, let us calculate the transistor power dissipation when no signal is applied, which is
=
VCEQ
x ICQ
=
8V
1.21A
=
lOW:::: Watts.
x
The power dissipation when the rated power is delivered
=
lOW - 6.66W
=4W
4.11
Heat Sinks
The purpose of heat sinks is to keep the operating temperature of the transistor low, to prevent
thermal breakdown. Due to increase in temperature, Ico increases. Due to increase in Ico' Ic
increases and hence power dissipation increases. Due to this, temperature increases and thus it is a
cummulative process. Due to this, the transistor will fail or breakdown occurs. To prevent this, heat
sinks are used to dissipate power to the surroundings and keep the temperature low.
The heat is transferred from the die to the surface of the package or casing of the ambient by
convention, from the surface to the ambient by convention and radiation. If heat sink is used, the
heat is transferred from the package to heat sink and from heat sink to the ambient. Heat sink
expedites the power dissipation and prevents breakdown of the device.
The rise in temperature due to power dissipation is expressed as Thermal Resistance expressed
in °C/w, and is symbolically represented as 9. It is the rise in temperature in °c due to 1W of power
dissipation. The equations governing this are,
9ja = 9jc + 9cn + 9na
9jc
=
(Tj - Tc) / P
9cs = (Tc - Ts) / P
9 sa = (Ts- Ta) / P
Power Amplifzers
181
9ja = Junction to ambient thermal resistance
9jc = Junction to casing thermal resistance
9cs = Casing to heat sink thermal resistance
9sa = Heat sink to ambient thermal resistance
Tj =
Tc =
Average junction temperature
Average case temperature
Tsa = Average heat sink temperature
T a = Ambient temperature
P=
Power dissipated in Watts.
Example: 4.5
What is the junction to ambient thermlll resistance for a device dissipating 600 m W into an ambient of
60°C and operating at a junction temperature of 120°C.
Solution:
Here heat sink is not considered.
9ja --
or
9.
=
T·-T.c
To
T· -Tc+T -T
T.-T
P
P
J
Ja
9ja =
T.
J
c- a
---=--+ -!!..-=
P
P
c
a = J
a
120-60 60
0.6 = -0.-6
= 100 °CIW
Fot- Transistor devices, the heat sinks are broadly classified as :
1. Low Power Transistor Type.
2. High Power Transistor Type.
Low Power Transistors can be mounted directly on the metal chassis to increase the heat
dissipation capability. The casing of the transistor must be insulated from the metal chassis to
prevent shorting.
Beryllium oxide insulating washers are used for insulating casing from the chassis. They have
good thermal conduetivity.
Electronic Circuit Analysis ,
182
Zinc oxide film silicon compound between washer and chassis, improves the heat transfer
from the semiconductor device to case to the chassis.
High Power Transistor heat sinks.
re TO-3 and TO-66 types. These are diamond shaped. For power transistors, usually, the ease itself
in the collector convention and radiation is shown in Fig. 4.34. The thermal resistance of the heat
sinks will be typically 30C/W.
Fig. 4.34 Fin-type heat sink
Fig. 4.35 Power transistor heat sink
Power Amplifzers
I. Amplifiers are classified based on
(a)
(b)
183
(c)
(d)
2. If the magnitude of signal is small and operating point swing is within the active region, that
amplifier is classified as _ _ _ _ __
3. Different types of coupling employed in amplifier circuits are
(a)
(b)
(c)
(d)
4.
Due to the input signal swing, if the operating point shifts into cut off and saturation regions,
that amplifier is classified as _ _ _ _ __
5. Conduction angles of large signal amplifiers are _ _ _ _ _ _ _ __
(b)ClassB _ _ _ __
(a) Class A
(c)ClassAB
(d) Class C _ _ _ __
6.
In class A power amplifiers the operating point Q is in _ _ _ _ _ _ of dynamic transfer
curve of the active device.
7.
In class B amplifiers, the Q point is set _ _ _ _ __
8. In class C amplifiers the operating point is set _ _ _ _ _~.
9.
Maximum theoretical efficiency of series fed amplifiers is _ _ _ _ __
10. Maximum effiCiency of transformer coupled amplifiers is _ _ _ _ __
II. The frequency range in which transformer coupled amplifiers are used is _ _ _ _ __
12. The maximum theoretical efficiency of class B push pull amplifier is _ _ _ _ __
13. For mirror symmetry or half wave symmetry, the mathematical equation is _ _ _ _ __
14. In push pull configuration, type of harmonics eliminated are _ _ _ _ __
15. In class B amplifiers, relation between maximum collector power dissipation Pc (Max) and
maximum output power dissipation Po (Max) is _ _ _ _ __
16. Cross over distortion occurs because of _ _ _ _ _ _ characteristic of E - B junction of
the transistors.
-17. Transformerless class B power amplifier circuit is _ _ _ _ _ _'
18. Complimentary symmetry circuit is so named because _ _ _ _ __
19. The complimentary symmetry circuit with single d.c. bias supply circuit is also
called
------
20. Phase Inverter circuits are also called - - - - - -
Electronic Circuit AnalyslI!I
184
21. What is the mode of operation of a last stage in a cascade?
22. Derive an expression for second harmonic distortion interms of Imax' Imm, Ie'
23. What is the expression for total harmonic distortion interms of second, third harmomic
24. What is impedance matching?
25. Why do we go for transformer coupled power amplifier?
-
26. What is the equivalent load resistance of a transformer coupled amplifier interms of turms
ratio?
.
27. Is the equivalent load resistance increasing or decreasing if n greaterthan 1 ?
28. What is conversion efficiency?
"'.
29. What is the lriiXimum \value of efficiency for the series fed load?
30. What is the maximum value of efficiency for a transformer coupled load?
31. How will the input signals be in a push pull amplifier?
32. What are the advantages of a push pull configuration?
33. Draw the waveforms to explain the class B operation.
34. What is the maximum efficiency of a class B amplifier?
Power Ampiifzers
185
1.
What are the different methods of clarifying electronic amplifiers? How arethey classified,
based on the type of coupling? Explain.
2.
Compare the characteristic features of Direct coupled, resistive capacitor coupled, and
Transformer coupled amplifiers.
3.
Distinguish between small signal and large signal amplifiers. How are the power amplifiers
classified ? Describe their characteristics.
4.
Derive the general expression for the ouput power in the case of a class A power amplifier.
Draw the circuit and explain the movement of operating point on the load line for a given
input signal.
5.
Derive the expressions for maximum. Theoretical efficiencY'for maximum.
(i) Transformer coupled
(ii) Serves fed amplifier what are thier advantages and disadvantages.
6.
Show that in the case of a class A transforms cgupled amplifier, with inpedence matching,
the expression for voltage gain AV is given as
hfe ) RL Nt
Av = - ( . -h. . -N with usual notation
2
Ie
2
7.
What are the advantages and disadvantages of transformer coupling?
8.
Show that class B push pull amplifiers exhibit halfwave symmetry.
9.
Derive the expression for Max. Theoretical efficiency in the case of class B push pull
amplifier. Why is it named so ? What are its advantages and disadvantages?
10.
Draw the circuit for composite tune amplifiers and explain its operation.
11.
What are phase inverter circuits? Draw a typical circuit and explain its working.
12.
Draw the pentode pushpull amplifier and explain its operation.
• 13.
Explain about Class D and Class S power amplifiers. Mention their sallent features and
applications.
14.
How are the tuned amplifiers classified? Explain the· salient features of each one
~~~
15.
!
Draw the circuit for single tuned capacitance coupled amplifier explain its operation.
Electronic Circuit Analysis
186
(b) Type of coupling
1. (a) Frequency range
(c) Output - power/conduction angle
(d) Magnitude of signal.
2.
Small signal amplifier
3.
(a) Direct coupling
(b) R - C coupling
(c) Transformer coupling
(d) L - C tuned coupling (e) series fed
4. Large signal amplifier
5.
(a) Class A 360
0
(c) Class AB ISO to 360
6.
(b) Class B IS0
0
0
(d) Class C < IS0
0
The centre of linear region of the
7. Near cut off of the active device.
S.
Beyond cut off
9. 25 %
10. 50 %
11. Audio frequency range 20 Hzs to 20 KHzs.
12. 7S.5 %
13. i (wt) = - i (wt + 1t)
14. Even Harmonics
15. Pc (Max) = 0.4 Po (Max)
16. Cut in voltage or threshold voltage .
.17. Complimentary symmetry circuit.
IS. Both PNP and NPN transistors are used.
19. Totempole circuit
20. Paraphase amplifiers.
21. Since for the last stage, the input signal has a high amplitude, the mode of operation will be
other than class A.
22. Refer to the derivation.
Power Amplifiers
23. TD =
187
~D; +D; +D~ + .....
24. For maximum power transfer to the load, the load impedance should be conjugate of the
effective impedance.
25. For impedance matching and hence higher efficiency.
I
26. RL
1
RL where n is the turns ratio.
n
= -2
27. R{ decreases
Signal power delivered to the load
28. 11 = - - " - - = - - - - - - - - - dc power absorbed.
29. 11 = 25%
30. 11
=
50%
31. The input signals are both 1800 out of phase.
32. Less harmonic distortion, More efficiency, Ripples in power supply are reduced, Magnetic
effects are reduced.
33.
o
--+ Vile
21t
34. 78.5%
"This page is Intentionally Left Blank"
UNIT - 5
Tuned Amplifiers - I
In this Unit,
• Different types of Tuned amplifier circuits are analyzed.
• Equivalent circuits of the output stages are given.
• FET Tuned R.F. amplifier circuits, wideband amplifier circuits, shunt
compensation aspects are also explained.
190
Electronic Circuit Analysis
5.1
Introduction
A tuned amplifier is one, which uses a parallel tuned circuit, as its load impedance. A parallel tuned
circuit, is also known as anti resonant circuit. The characteristics of such an anti resonant circuit is
that its IZI is high, at the resonant frequency, and falls off sharply as the frequency departs from the
resonant frequency. So the gain versus frequency characteristics of a tuned amplifier will also be
similar to the IZI characterisr~s of the resonant circuit. When IZI is maximum, V(} will also be
maximum. This is for AC. IZI ~S\considered.
;~:J\~"
"
jZj
i
(load)
~f
~f
Fig. 5.1 Tuned amplifier characteristics
5.1.1
Applications
Tuned amplifiers are used to amplify a single radio frequency or narrow band of frequencies ..
So basically they are used in 1. RF amplifiers 2. Communication receivers
Tuned amplifiers use variable Inductance (L) or variable Capacitance (C) to vary the resonant
frequency.
In tuned amplifiers, harmonic distortion is very small, because the gain of the amplifiers is
negligibly small for frequencies other than to (the resonant frequency). So Harmonics which on
of higher frequencies will have very low gain and hence harmonic distortion will be less for
tuned amplifiers.
5.1.2
Classification
Tuned Amplifiers
I
I
j
Large Signal
Small Signal
*
To amplify low RF signals
*
To amplify large RF signals
*
*
Power output is low
*
Power output is more
Operated in class A
*
Operated in class B, class C or
class AB modes.
*
Pushpull configuration used to
further
reduce
harmonic
distortion.
I
Single tuned
Double tuned
Staggered tuned
This classification is similar to the classification ofpower amplifiers.
Tuned Amplifiers - I
191
Small Signal Tuned Amplifiers
I
Single Tuned Amplifiers
Double Tuned Amplifiers
Staggered Tuned Amplifiers
5.1.3 Single Tuned Amplifier
Uses one parallel tuned circuit as the load IZI in each stage and all these tuned circuits in different
stages are tuned to the same frequency. To get large Av or Ap , multistage amplifiers are used. But
each stage is tuned to the same frequency, one tuned circuit in ~me stage.
5.1.4 Double Tuned Amplifier
It uses two inductively coupled tuned circuits, for each stage of the amplifier. BotQ the tuned circuits
are tuned to the same frequency, two tuned circuits in one stage, to get sharp response.
5.1.5 Stagger Tuned Amplifier
This circuit uses number of single tuned stages in cascade. The successive tuned circuits are tuned
to slightly different frequencies.
Single tuned amplifiers are further classified as :
I
I
5.2
I
Transformer coupled or inductive coupled
Capacitive coupled
Single Tuned Capacitive Coupled Amplifier
L, C tuned circuit is not connected between collector and ground because, the transistor will be short
circuited at some frequency other than resonant frequency.
vs
~
Fig. 5.2 Single tuned capacitive coupled amplifier
The output of the tuned circuit is coupled to the next stage or output device, through capacitor
Cb. So this circuit is called single tuned capacitbe coupled amplifier.
R I , R2, RE, CE are biasing resistors and capacitors. The tuned circuit formed by Inductance
(L) and capacitor (C) resonates at the frequency of operation. .
Transistor hybrid 1t equivalent circuit must be used since the transistor is operated at high
frequencies. Tuned circuits are high frequency circuits.
R j = input resistance of the next stage.
192
Electronic Circuit Analysis
R,
E
Next stage
Fig. 5.3 Equivalent circuit
Modified equivalent circuit using Miller's Theorem.
According to Miller's theorem, the feedback capacitance Cc is Cc (1 - A) on the input side
and Cc
(A~ I) on the output side. But where as resistance is (I ~.~)
on the input side ( ; ; I)
on the output side.
~ B
fbb'
C'
E
Fig. 5.4 Equivalent circuit (applying Miller's Theorem)
The equivalent circuit after simplification, neglecting (
B
~)
B'
is shown in Fig. 5.5.
C
1
R=o hoe
Fig. 5.5 Simplified equivalent circuit
Y
1
=
I
=
R + jroL
Input admittance as seen by II stage.
Instead of L and R being in series, they are being represented as equivalent shunt
Rp and Lp for parallel
I
I
Rp
jroLp
=-+--
element~
Tuned Amplifiers - I
193
where
Inductor is represented by Rp in series with inductance Lp.
Q at resonance,
roOL
Qo=T
roL» R
.: Resistance of the inductor R is small,
Lp=L
neglecting R2 compared to ro 2 L2.
Therefore output circuit is simplified to,
c
R.I
Fig. 5.6 Simplified circuit
1
1
1
1
Rt
R
Rp
Ri
-=-+-+1
ro = - o JLC
~
is the input Fesistance of the next stage
Qe is defined as,
~ =
Q=
e
Suspectance of L or capacitance of C
Conductance of shunt resistance R t
-~-------~--------
resistance of tuned circuit
Qe=roOC~
=
~ tllroOL) = (roo C]
roo L
ljRt
IjRt
Electronic Circuit Analysis
194
Let roo be the resonant angular frequency in rad/sec.
1
roo
.JLc
=
Output voltage Vo = -gm Vb'e·Z (-gmVb'e is the current source).
where Z is the impedance of C, Land R t in parallel.
Y
Admittance
= -
1
Z
1
I
= - + - - + jroC
R
t
jroL
Multiplying by Rr throughout and dividing,
or
Y= _1_
[1 +~+
jroCR 1
R
jroL
t
t
=
1 [
. roo roC R t
R t roo
-R 1+ J
+ .
L
t
roo
JroO ro
1
(Multiplying and dividing by roo)
Q = e
where
Rt
w
o
L
=ro o'~
CU)
Qe = roo C Rt
=~
roo L
z
l+jQ
Susceptance of L or C
Conductance of shunt resistance R t
Qe is defined as
Let
i.e.,
0
1
ro -roo
e [ roo
ro
=
Fractional frequency variation
variation in frequency expressed as a fraction of the resonant frequency
195
Tuned Amplifiers - I
Rewriting the expression for Z, as
1+0
R
t
Z-------=:-----=
- l+j2Qe o[I+0/2]
1+0
If the frequency ro is close to resonant frequency roo' 0 «
Therefore Simplified expression for Z is
Z
At resonance,
At resonance,
~
Rt
= -----''---
1+ j2Q e o
ro = roo' 0 = 0
may also be put as,
Rp= Q02 R=Q o =
Expression for
~
rb'e
Vb'e = Vi .
=rooLQo
potential divider network
rbb' + rb'e
Expression for
Vo = -gm Vbe . Z
Expression for
Voltage gain
1.
Vo
rb'e
= -gm·
.Z
A = Vi
rb'e + rbb'
A =-gm.
fb'e
Rt
fb'e +fbb' 1+ j2BQ e
voltage gain at resonance. Since at resonance 0 = 0
-gm·rb'e
.~
rb'e +fbb'
A
--=---A
reso
Electronic Circuit Analysis
196
A
Magnitude
Areso
A
Phase angle
= -
tan-I (2 0 Qe)
Areso
At a frequency
0)1'
below the resonant frequency 0 has the value,
I
=-2Q ;
e
A
1
- - = - =0.707
Areso
ro l is the lower 3db frequency.
Ji
Similarly ro 2, the upper 3db frequency is
o=+
The 3 db band width Aro
=
1
A
1
2Q ; Areso =
e
(ro 2 - ro l )
Ji
= 0.707
Kro 2 - ro O)+ tro O - rol)J· roo
roo
=
But
5.3
[0 + 0] roo
=
2 0 roo
1
0-2Q
e
Tapped Single Tuned Capacitance Coupled Amplifier
II Stage
Fig. 5.7 Tapped single tuned capacitive coupled amplifier circuit
197
Tuned Amplifiers - I
5.3.1
Equivalent Circuit on the Output Side of the I Stage
RI is the input resistance of the II stage.
Ro is the output resistance of the I stage amplifier.
II
Fig. 5.8 Equivalent circuit
The input IZI of the common emitter amplifier circuits will be less. So the output impedance of
the circuit being coupled to one common emitter amplifier, should also have low IZI for impedance
matching and to get maximum power transfer. So in order to reduce the impedance of the LC
resonant circuit, to match the low IZI of the common emitter circuit, tapping is made in the LC tuned
circuit. Tapped single tuned circuits are us~d in such applications.
5.3.2
Expression for 'Inductance' for Maximum Power Transfer
Let the tapping point divide the impedance into two parts LI and L2.
Let LI = nL so that L2 = (1 - n)
Writing Kirchoff's Voltage Law (KVL)
VI = jroL . II - jro (L2 + M) 12
..... (1)
o
..... (2)
= -jro (LI
+ M) II + (Rj + jroL2) 12
Where M is the mutual inductance between LI and L2. Solving equations 1 and 2,
I
=
I
Hence the
VI {Ri + jroL 2 )
2
jroL (Ri + jroL 2
(L2 + M
---:----"-"--:--..,...:::--~
)+00
f
..... (3)
IZI offered by the coil along with input resistance ~i of the next stage is
..... (4)
..... (5)
But roL2 much less than Ri .
As Ri, the input resistance of transistor circuit II stage is KQ and much greater than roL2
..... (6)
Electronic Circuit Analysis
198
M
=K~LI L2
M
=
Mutual Inductance
Where K is the coefficient of coupling. Since Ll
=
nL, L2
=
(l-n)L
..... (7)
Putting K = I, we get
M::: L
~n-n2
..... (8)
Substituting thus value ofM in (6),
..... (9)
..... (10)
GcoL+ R)
The resistance effectively reflected in series with the coil due to the resistance R, is given by,
This is the resistance component;
$: series, i : input
This resistance R is in series with the coil L may be equated to a resistance Rip in shunt with the
coil where Rip is given by, ~
Rip=
(coLi /
Ris
So the equivalent circuit is
,'-
c
Source
R
----)~
Fig. 5.9 Equivalent circuit
load
'P
Tuned Amplifiers - I
199
Simplifying,
Fig. 5.10 Equivalent circuit after simplification
1
1
1
1
R tt
RO
Rp
Rip
-=-+-+--
Rtt
Qe = ro L
o
tt : tuned tapped circuit.
1
roo
=
JLC
Under the conditions of maximum power transfer theorem, the total resistance appearing in
shunt with the coil is = Rop
Since it is a resonant circuit, at resonance, the IZI in purely resistive. For maximum power
transfer IZI = R!2.
But
Solving for L, we get
RO tQ O -2Qe)
2roO QoQ e
Expression for L for maximum power transfer.
L
=
---'''--~---
L~~[~-<tl
This is the value of L for maximum power transfer.
~
Electronic Circuit Analysis
200
Expression for voltage gain and Bandwidth are determined in the same way as done for a single
tuned circuit. In this circuit we have,
1. Ru instead of ~ (as in single tuned) tapped tuned circuit.
2. Output voltage equals (l - n) times the voltage developed across the complete coil.
IZI at any frequency close to Wo is given by,
Z
output voltage
=
Vo=
R tt
1 + j 28 Q
Ru =
e
-gm Vi rb'e
rb'e + rbb'
resistance of Tapped Tuned Circuit
.Z(1-n)
.. voltage gain
rb'e
-gm (1 - n). --=-=-rb'e + rbb'
At resonance, voltage gain is
=
A
1
=----Areso
1- j 28Q e
5.4
Single Tuned Transformer Coupled or Inductively Coupled Amplifier
C'
z
(a)
I II stage
MO~
~----~----~~~~
~
(b) Equivalent output circuit
Fig. 5.11 Inductive coupled amplifier circuit (a) and its equivalent (b)
Tuned Amplifiers - I
201'
in this circuit, the voltage developed across the tuned circuit is inductively coupled to the next
stage. Coil L, of the tuned circuit, and the inductor coupling the voltage to the II stage, L2 form' a
transformer with mutual coupling M. This type of circuit is also used, where the input IZI of the
II
stage is smaller or different from the tuned circuit. SO IZI matching is done by the transformer
depending on its tum ratio. In such requirements, this type of circuit is used.
The resistors R I, R2 and R; and R2 are the biasing resistors. The parallel tuned circuit,
Land C resonates at the frequency of operation. Fig. (b) shows output equivalent circuit. Input
equivalent circuit will be the same as that of the capacitive coupled circuit.
In the output equivalent circuit, C is the total capacitance, including the stray
~ 1). L2 and R2 are the inductance and resistance of
capacitance, Miller equivalent capacitance C ( A
the secondary winding.
5.4.1
Expression for Ll for Maximum Power Transformer
Writing KVL to the primary and secondary windings,
V = II Zll + 12 Z12
o = II
where
Z21 + 12 Z22
ZII = R + jroL
ZI2 = Z21 = jroM
Z22 = R2 + Rj + jroL2
Solving eqs. (1) and (2) for II'
1
I
V,Z22
--~=--::--
=
..... (1)
..... (2)
..... (3)
..... (4)
..... (5)
..... (6)
2
ZII Z22 -Z12
The impedance seen looking into the primary is,
Zjn
=
=
2
V Zl1 Z22 - Z12
11
Z22
2
Z12
Zll - - Z22
..... (7)
Substituting the values of Zll' Z22 and ZI2 in equation (7) we get,
ro 2 M2
Zjn = (R + jroL) + R
R
. L
2 + i + Jro 2
..... (8)
R j generally much greater than R2 and roL2.
ro 2 M2
Zin::: (R + jroL) + - - -
R.1
..... (9)
Electronic Circuit Analysis
202
L
c
Fig. 5.12 Equivalent circuit
(02 M2
- - - is the impedance of the secondary side reflected to the primary.
R.
I
IfM is reasonably large, then R«
(02 M2
---
R.1
(02 M2
Zin :::: j(OL + - - R.
...... (10)
I
The equivalent circuit may be written as,
(02M2
Inductance L with series resistance
R.1
may be represented as L in shunt with
Rio as
shown below, where
..... (11)
c
L
RIp
Fig. 5.13 Simplified circuit
For maximum transfer of power at resonance,
Rip
=
Ro=
Ro
(~r·Ri
Equation 12. gives the value of M for maximum power transfer.
..... (12)
Tuned Amplifiers - I
203
Fig. 5.14 Equivalent circuit
L and L2 are the primary and secondary windings of inductances.
M=K~LL2
..... (13)
Combining eqs. (12) and (13) we get
..... (14)
Therefore from equation 14, for a given value of Ro and coefficient of coupling K and ~, we
can determine L2 for maximum transformer of power.
Shunt resistance
Ro and Rip may be combined to yield the total shunt resistance Ru.
1
1
1
R tt
RO
Rip
-=-+-R tt = Resistance of tapped tuned circuit
..... (15)
Effective Q of the entire circuit is,
•
R
Q = _tt_
e
co L
....• (16)
o
where Wo is the resonaqt frequency of Land C.
1
COo =
.JLC
..... (17)
Under conditions of maximum transfer of power, total resistance appearing in shunt with the
coil equals Ro/2. Since it is respnant circuit, at resonance,
IZI = resistance only.
For
maximum power, R = Rl2.
equation 16 becomes
Ro/2
o
Qe = -;--L
or
Ro = 2 Qe COo L
..., I..,{lar
..
..... (19)
Electronic Circuit Analysu.
204
Solving equations 1 and 2
..... (20)
..... (21)
..... (22)
IZI of the output circuit at any frequency 'ro' close to roo'
Impedance of output circuit Z
=
is given by,
R
1 + j 2~ Q
e
..... (23)
..... (24)
..... (25)
.. Voltage gain A at any frequency ro is,
R i Z 21
Rtt
Zll Z22 - Zf2 . 1+ j2&Qe
..... (26)
Voltage at resonance,
A
--=----
5.5
..... (27)
CE Double Tuned Amplifier
In this circuit, voltage developed across the tuned circuit in the collector circuit is inductively coupled
to another tuned circuit. Both circuits are tuned to the same frequency, the desired frequency of the
input signal.
Tuned Amplifiers - I
5.5.1
205
Advantages
1. It provides larger 3-db band width than the single tuned amplifier. Therefore
Gain x Bandwidth product is more. (eventhough gain is same, Bandwidth is more.
So gain x Bandwidth product is more).
2. It provides gain - frequency curve having steeper sides and flatter top.
"
In double tuned amplifier, as Bandwidth is increased the overshoot of gain also increases.
So a compromise must be made. The value of the coefficient 'b' ranges from 1.0 to 1.7.
5.5.2
Circuit
Fig. 5.15 Double tuned amplifier circuit
5.5.3
Equivalent Circuit
c
L
RI
R
Fig. 5.16 Equivalent circuit
Voltage developed across the tuned circuit in the LCR combination is inductively coupled to another
tuned circuit of the same resonant frequency.
R is the resistance associated with L. Rj is the output IZI of the second circuit. The equivalent
circuit can be further simplified as, Rj and R2 are combined to form R j '. Ro in parallel with L-R may
be brought in series with inductance and combined with R to from R i .
The current source gmV b'e in shunt with
with 'C'.
'c'
is modified to a voltage source VI in series
:. The modified circuit is as shown in Fig. 5.17.
Electronic Circuit Analy~is
206
RI=R
+R2
j
j
Zon
Rjl=RolIl\
Fig. 5.17 Modified circuit
ro 2 M2
= --~---"7"
Z·
Rik +L 2 -
m
At resonance,
and
1
roo L = ro C
..... (2)
o
roo L2
=
roo
C
.,U
..... (1)
..... (3)
2
.. At resonance, Zi is reduced to
Zin
=
ro 2 M2
--=-0__
R!1
..... (4)
Hence at resonance for maximum transfer of power Ro must equal Zin' i.e., M is adjusted to
critical value Me, such that
ro 2 M2
Ro= oR. C
...... (5)
M
= roo Mo
..... (6)
=roo Kc~LL2
..... (7)
1
or
RO Ri
where Ke is the critical value ofthe coefficient ofcoupling corresponding to the critical value
Me of mutual inductance.
From 7,
.. ... (8)
...... (9)
107
Tuned Amplifiers - I
In a general case,
So let
K =b.
Now,
VI = Zll II + Z12.1 2
K:;t: Ke
Kc
..... (10)
..... (11)
Solving equation 10 and 11, for 12, we get
VI Z2I
12 =
2
..... (12)
ZII Z22 -Z12
where
Zll
=
1
,
Ro + j (roL - roC)
~2
=
1 )
Ri, + J. ( roL2 - ~
and
ZI2 = ~I = jroo M
At resonance,
Zll = R~ and ~2 = Ri
...... (13)
..... (14)
..... (15)
For maximum transfer of power at resonance,
Zll = R~, ~2 = Ri,
..... (16)
ZI2 = jroo Me
Substituting these values of Zll' ~2' and ZI2 in equation 12, we get the following conditions
for maximum transfer of power at resonance.
-j VI roOMC
12 Max = RI RI
2 M2
o i +ror C
..... (17)
Me is the critical value of Mutual Inductance.
Substituting the value of Me in equation 17, we get
..... (18)
Electronic Circuit Analysis
208
Magnitude
(j)(j)o -
where
b
K
=
Kc
=
~
±
~( b2 -I ± 2b)/Q
(KIKc)
Coefficient of coupling
Critical value of coefficient of coupling corresponding to
the critical value of Mutual Inductance Mc.
Tuned amplifiers are used where it is desired to amplifY a relatively narrow band offrequencies,
centered about some designated mean carrier frequency.
5.6
Applications of Tuned Amplifiers
I.
2.
3.
4.
Radar
Television
Communication receivers
I.F amplifiers
There are mainly three types tuned voltage amplifiers.
1. Single tuned amplifier
2. Double tuned amplifier
3. Stagger tuned amplifier
209
Tuned Amplifiers - I
1.
Radio Frequency Range is _ _ _ _ _ _ _ __
2.
VHF Frequency Range is _ _ _ _ _ _ _ __
3. Parallel Tuned Circuit is also known as
---------
4.
Generally Tuned Amplifiers are used in _ _ _ _ _ _ _ __
5.
In Tuned Amplifiers, Harmonic Distortion is _ _ _ _ _ _ _ __
6.
Small signal Tuned Amplifiers are operated in _ _ _ _ _ _ _ _ _ mode.
7.
Large signal Tuned Amplifiers are operated in
8.
Small signal Tuned Amplifiers are classified as
(a) _ _ _ _ _ __
9.
modes.
(b) _ _ _ _ _ __ (c) _ _ _ _~--
Single Tuned Amplifiers are classified as _ _ _ _ _ _ _ __
10. In Tuned Amplifiers equivalent circuits, the model used for Transistors (BJT) is
II. Expression for
~
for tapped single tuned capacitance coupled amplifier with usual notation
Areso
is
---------
12. Double Tuned Amplifier provides _ _ _ _ _ _ _ _ _ Bandwidth than single tuned amplifier.
13. Common Emitter (C.E) Double Tuned amplifier provides Gain - Frequency curve with
sides
.
top.
---------
14. In Radars and Television _ _ _ _ _ _ _ _ _ type of amplifiers are used.
210
Electronic Circuit Analysis
I. (a)
(b)
(c)
What are the different types of Tuned Amplifiers?
How are Tuned amplifiers classified?
What are the applications of Tuned amplifiers?
2.
Draw the circuit for single tuned capacitive coupled amplifier and explain its working. Draw its
equivalent circuit and derive the expression for (AIAreso)
3.
Draw the circuit for tapped single tuned capacitance coupled amplifier and explaiI1 its working.
Draw the frequency response. Derive the expression for L for maximum power transfer.
4.
Draw the circuit for single tuned inductively coupled amplifier. Draw its equivalent circuit and
derive the expression for (A:
).
so
s.
Draw the circuit for Double Tuned Amplifier. Explain its working. What are the advantages of
this amplifier? Derive the expression for 12 Max.
Tuned Amplifiers - I
211
1.
>20 KHzs
2.
30 MHzs - 300 MHzs
3.
4.
5.
6.
Anti Resonant Circuit
Class A
7.
Class B, class C or class AB
B.
(a) Single Tunded
(b) Double Tuned
9.
(a) Capacitive coupled
(b) Transformer coupled or Inductive coupled
10.
Hybrid - 1t equivalent circuit
11.
- - = - - - - where (5 =
R. F amplifiers, communication Receivers
Less
A
Areso
1+ j2(5Qe
12.
Larger
13 .
Steaper sides and flatter top
14.
Tuned amplifiers
[ro-roo)
"'0
VJ
(c) Staggered Tuned
R
t
Q=roCR=-.
e
0
t
roL
o
"This page is Intentionally Left Blank"
UNIT - 6
Tuned Amplifiers - II
In this Unit,
• Single stage amplifiers in the three configurations of C.E, C.B, C.C,
with design aspects are given.
• Using the design formulae for Avo AI' R Ro etc, the design of single
"
stage amplifier circuits is to be studied.
• Single stage JFET amplifiers in C.D, C.S and C.G configurations are
also given.
• The Hybrid - 1t equivalent circuit of BJT, expressions for Transistor
conductances and capacitances are derived.
• Miller's theorem, definitions forlp and!,. are also given.
• Numerical examples, with design emphasis are given.
214
6.1
Electronic Circuit Analysis
Stagger Tuning
Tuned amplifiers have large gain, since at resonance, Z is maximum. So Av is maximum. To get this
large Av over a wide range of frequencies, stagger tuned amplifiers are employed. This is done by
taking two single tuned circuits of a certain Bandwidth, and displacing or staggering their resonance
peaks by an amount equal to their Bandwidth. The resultant staggered pair will have a Bandwidth,
..fi times as great as that of each of individual pairs.
~f
Fig. 6.1 Variation of
6.2
Av with f
Single Tuned Transistor Amplifier
Va.;
Fig. 6.2 Single tuned amplifier
6.3
Stability Considerations
An electronic amplifier circuit or oscillator circuit becomes unstable, i.e;, will not perform the desired
function, due to various reasons associated with circuit design aspects like Thermal stability, Bias
considerations, output circuit, feedback, circuit etc. Let us consider th~se aspects.
Thermal Effects: Any electronic circuit, when used continously, various components will get heated,
due to power dissipation in each of the components. When the active..device in the circuit (BJT) also
gets heated, due to the dependance of its characteristics on temperature, the operating point changes.
So the device will not function as desired and the output from the eiectronic circuit will not be obtained
as per specifications. Heat sinks are to be used, to dissipate the excess thermal energy and to keep the
operating temperature of the active device within the limits. Heat sinks are good themal conductors
designed suitably, to keep the temperature of the device within limits.
21S
Tuned Amplifiers - II
Let
PJ = Power dissipated in watts at the junction of the device
T A = Ambient Temperature
TJ
=
Temperature of the junction
0T = Themel resistance in °C/W
TJ = TA + 0T' PJ
Range of value of OJ is 0.2 to 2.0 °C/W.
A thin mica spacer is often used to electrically insulate the BJT or semiconductor device from the
heat sink. The thermal path for heat dissipation is from semiconductor p-n junction to casing of the
device, from the case to heat sink and from heat sink to the ambient.
°c
= Thermal resistance in °C/W associated with the Casing ofthe device
0H = Thermal resistance in °C/W associated with the Heat sink.
OR = Radiation Thermal Resistance
So,
TJ - PJ (OJ + 0c)
0c OR
Total Heat sink resistance = 0H +
C+ R
TH
=
° °
If PJ ' the power dissipated at the junction is independent of TJ
then,
dPJ =_1
dTJ
aT
dP
1
At the operating temperature, - J > dTJ
aT
Bias Considerations: Distortion in Audio amplifiers and other types of circuits depends on :
(i) Input signal level (in mv)
(ii) Source Resistance
(iii) Bias Conditions
(iv) Type of output load and its impedance
(v) Loading effect.
For audio frequency amplifiers, distortion of I or 2% is considered maximum for higher fiedility
equipment.
A maximum of 5% distorion is set sometimes for radio receivers and public address systems.
Electronic Circuit Analysis
216
Consider the circuit shown in Fig. 6.3.
R'
L
b
Fig. 6.3
Expression for input power Psis,
'
Ps = Ib2 (RS' + hie)
Expression for output power Po is,
Po
=
Ic 2 .R L'
IC 2R L '
.'. Power Gain
2
Ib (Rs' + hie)
hf/R L '
= ----"=---=-
Rs' + hIe
Considering only the BJT, expression for power gain
A p(8JT) =
hf/R L
h.
Ie
= hre' ~ RL '
Maximum power is transferred to the base-emitter circuit of BJT, when impedance matching is
done as,
Rs'
= hIe
:. Matched input Power Gain
Ap (Matched input) =
hfe 2 RL '
2h.
Ie
- b
gm·RL'
Ap- fe'
2
To maintain output voltage and power levels stable, the circuit must be designed accordingly.
6.4
Tuned Class B and Class C Amplifiers
The efficiency of the output circuit of an amplifier increases as the operation is shifted from class A
to B and then to C.
In class C amplifiers, efficiency approaches 100%.
217
Tuned Amplifiers - II
But the difficulty with class 'C' operations is harmonic distortion is more. So the conventional
untuned amplifiers use class B pushpull configuration. But, if it is a tuned amplifier, and only one
frequency fo' is to be amplified and power to be handled Po is large, then class C operation is
preferable, since efficiency is high and harmonic distortion will not be a problem since, only one
frequency is to be amplified and the tuned circuit will reject the other frequencies.
tv
0.707 Av
0.707 Av
-+
f
Fig. 6.4 Frequency response
In radio transmitters, the output power often exceeds 50 KW and efficiency is the important
criteria. Hence usually class C operation is done. The operation of such amplifier circuits is non
linear, since the variation of output current iL with the input current ib is non linear, as the signal
magnitudes are large. So these amplifiers are also called as Large signal amplifiers.
Since they have a tuned circuit and usually operate in the R.F range, they are also called as
R.F power amplifiers.
Usually these circuits use common emitter configuration, but sometimes common base
configuration is also used. A simplified circuit (without emitter stabilization) uses transformer and
FET as shown below. Since the circuit is class B and class C configurations, the input signal itself
drives the transistor or FET into conduction. So there are no separate biasing resistors.
The circuit may be class B or class C, depending upon the actual conduction angle (180 0 or less).
6.4.1
Bipolar Junction Transistor (BJT) Tuned Class BIC amplifier
The capacitors Cs , Cb and Cc are considered to be short circuits at the operating signal frequencies.
~i
Fig. 6.5 Tuned class BIC amplifier
The transformer circuit L l , C l and Rb, Cb provide high input impedance IZI to the input signai
and so ib passes to the base of the transformer only and will not get divided across Ll C l or Rb Cb.
So the power output will be more.
218
6.4.2
Electronic Circuit Analysis
FET TImed R.F Amplifier
Fig. 6.6 FET tuned RF amplifier
6.4.3
Waveforms
Vo
t
---,--
Vooi
7tI2
1t
I
I
I 21t
I
I
I
I
I
37t12
io
t
-70lf
Olf
0)
°2
7tI2
1t
37t12
0)
°2
21t
Fig.6.7 Frequency response
6.4.4
Resonant circuit
In a tuned amplifier, the functions of the resonant circuit are :
1. To provide correct load impedance to the amplifier.
2. To reject unwanted harmonics
3. To couple the power to the load
The resonant circuits in tuned power amplifiers are sometimes called tank circuits, because the
Land C elements store energy like water stored in a tank. Such a circuit is as shown in Fig. 6.8.
Its equivalent is (R + jroL) in parallel with Rv
Fig. 6.8 Resonant circuit
Tuned Amplifiers - II
219
Neglecting the losses in the capacitor C, the 11 of the above circuit is,
Efficiency
=
11
=
output power across RL/lnput power
2
IL·R L
IL = current through load resistance RL
Ia = current through inductor inductance with its internal res.istance R in series.
At resonance, the
IZI
is only, resistive.
R = Series resistance of 'L'. RL I is internal resistance
Let
=
Q of the coil at resonance
Let
= Q of the coil shunted by R
L. (Taking RL into consideration)
This equation can be written as,
1
1
----
Efficiency
ef_f
= 11 = _Q--=e=ff_Q_O
-",- = 1- _Q_
IjQ eff
QO
Q o = Q factor at resonance frequency /0.
Qeff = Effective value of Q
Qo» Qeff' 11::::: 1 or 100% (or it is very high). IfRL can be varied, Qeffcan be made as small
as desired. The value of Qo is usually large.
Electronic Circuit Amuy:,.",
220
The purpose of resonant circuits in tuned circuits is,
I. To provide correct load
IZI
2. To reject unwanted harmonics
3. To couple power to load
These tuned circuits are sometimes called Tank circuits because, like water tank, energy can
be stored in inductance and capacitance energy storage elements, if they are ideal.
6.4.5
Tank Circuits
---7
I
Fig. 6.9 Tank circuits
RL = equivalent circuits RL in series with inductance, resistance
Neglecting losses in capacitors,
For the equivalent circuit,
For the equivalent circuit
I~
Rt,
--;-!::-----!"'--~
I~(R+RL)
Let
Qo =
mOL
R
=
= Q of the coil at resonance.
mOL
R +R1
. L
= Q of the coil shunted by RL·
Tuned Amplifiers - II
221
Q
eff
=\--
QO
6.4.6
Mutual Inductance Coupled Output Resonant Circuit
Fig. 6.10 Mutual Inductance coupled resonant circuit
6.4.7
Equivalent Circuit
Power dissipated in R2 and RL
Fig. 6.11 Equivalent circuit
R
Power dissipated in RL can be obtained by multiplying the above equation by R
6.5
LR
2+ L
Wideband Amplifiers
The frequency response of a given amplifier can be extended by adding few passive circuit elements
to the basic amplifier.
222
6.5.1
Electronic Circuit Analysis
Shunt Compensation
One simple method of shortening the rise time in the response of an amplifier circuit and thus enhancing
the high frequency response of the amplifier. (pulse having, sudden change in the input is considered
as high frequency variation) is to add an inductor in series with, Vcc and collector.
Fig. 6.12 Shunt compensation circuit
In the a.c equivalent circuit, this 'Inductance' will be in parallel with capacitor. Thus 'Inductance
Capacitor' combinations changes the output response. Since in shunt in the output stage, this is
called shunt compensated amplifier. The collector circuit Z = Rc + jwL.
This increases with frequency. So Vo increases and gain increases. So if 'Inductance' is not
present, the gain will be less. When 'capacitor' is present in the output circuit, Xc decreases as!
increases. Thus L will compensate for capacitor.
So for compensated amplifier, when L is introduced, the damping factor K is defined as,
K=Rc.~
1
fi=21tRC
c
Where fi is the upper 3 db point of the uncompensated amplifier.
~f
Fig. 6.13 Frequency response
So the frequency response changes, when 'L' is added:
The bandwidth of an amplifier circuit can be increased by decreasingfi and increasingfi.
223
Tuned Amplifiers - II
C
,..-_ _-, I
r-------,
I
C2
I
L _ _ _ _ _ _ _ .J
Fig. 6.14 Simplified circuit
12
r b,.
C
C2
gm Vb'.
EI o-----~--------------~--------~--~----_o E2
C = C. + CC(I-A) Where A is the gain of the angle (I stage)
Fig. 6.15 Equivalent circuit
6.5.2
Extension of Low Frequency Range
Low frequency range can be extended by decreasing II (Since the region as seen from the mid
frequency range becomes more iffi is decreased). For resistance capacitance coupled amplifier,
Where Cc is the coupling capacitor,jl can be decreased by increasing Ce , If the value of Cc
were to be large, its cost and size will be more. So there is a limit to which Cc can be increased.
Hence, to get lesser value of fl' R should be increased. This is known as Compensation. Instead of
changing the value ofC, to get decrease infi we are changing the value ofR. Thus R is compensating
for the effect of C.
Now if R were to be fixed and load Z were to change with frequency, another capacitor Cc is
connected in parallel with load resistance RL so that the net load Z, is ZL so that ZL increases as I
decrease, So fl can be increased. This is known as Compensation.
But the value of II is very small (few hundreds MHz) compared to ii. Therefore jro is
increased effectively by increasing/2 .
224
6.5.3
Electronic Circuit Analysis
Circuit for Extending Low Frequency Range (Fig. 6.16)
(a) Kinks in frequency response
6.5.4
Low Frequency Equivalent Circuit with Transistor Replaced by a Current Source
(b) Equivalent circuit with current source
Fig. 6.16
6.5.5
Extension of High Frequency Range
Bandwidth ofa given amplifiers can be effectively increased by increasing h. (sinceh» jj)h can
be increased by decreasing Cs . But there is a limit to which Cs value can be reduced h can also be
increased by decreasing RL 1. But if RL 1 is reduced the mid band gain will reduc~ Therefore without
changing the value of Cs and RL 1 and without decreasing the mid band gain, h can be increased by
compensating techniques.
There are two different methods:
(i) Series compensation
(ii) Shul'I:t compensation
This classification is depending upon character 'Inductance' is in series or in shunt with the
oad resistance in the a.c equivalent circuit.
where
m = 002L1R1 (R 1 is the resistance in series or the resistance of the coil itself)
is the upper cutoff frequency of the uncompensated amplifier.
00 2
225
Tuned Amplifiers - II
For a shunt peaked amplifier, for different values of m, the response will be as shown.
~f
Fig. 6.17 Kinks in frequency response
If m
= 0.414, ffi~ = 1.72
ffi2
where
ffi21
is the upper cutoff frequency with compensation.
For this value of 'm', there will not be peaking of the mid band gain. For higher value of m, there will
be peaking. So m = 0.414 value is often used. Because of res"nance, the voltage across Cs or thE
output voltage will be maximum. So overshoot occurs.
6.5.6
Series Peaked Circuit
L
Fig. 6.18 Series peaked circuit
C
Series peaked circuit is used if C
IC = a = 0.25.
1+ 2
If any other type of capacitances and shunt capacitances are present, this is not used. Shunt
compensated circuit is preferred over series peaked since, for shunt compensated value circuits, the
gain falls smoothly beyond fi. For series compensated circuits, there will be sudden drop in gain
beyondfi.
Because of the presence of both Inductance and Capacitance, in the circuit the fi point will be
near about the resonant frequency of the circuit. Overshoot or peaking in gain will occur near
about the resonant frequency. This will depend upon the Q factor of the circuit. As Q increase peak
gain increases.
6.6
Tuned Amplifiers
6.6.1
Impedance Transformation
When a signal generator is to supply power to a load or amplifier is to supply output power to a load
(driving a load), for maximum power transfer the output impedance of the signal generator or amplifier
Zo must be complex conjugate of load impedance ZL according to maximum power transfer theorem.
In other words, their magnitudes must be the same.
i.e.,
I ZL I = I Zo I
226
Electronic Circuit Analysis
This is konwn as Impedance Matching.
In the case of audio amplifiers, the load is usually a loudspeaker. Its impedance is of the order
of 40, 8Q or 16Q. But the audio amplifier circuit may not have output impedance equal to these
values. So impedance matching condition is not satisfied.
Similarly a radio transmitter will have typically output impedance of 40000. It is required to
supply power to an antenna. Antenna will hae usually impedance of 70Q. So impedance matching
condition is not satisfied for maximum power transfer, impedance transformation must be used in
such coupled circuits.
Some of the methods by which impedance transformation is done in coupoled circuits are :
1. Transformation of impedances with tapped resonant circuits
(a) by tapping inductors
(b) by tapping capacitors
2. Reactance section for impedance transformation
3. Image impedances - Reactance matching
4. Reactance T Networks for impedance transformation
6.6.2
Transformation of impedances with tapped resonant circuits
Consider a parallel LC tuned circuit connected to a generator. At antiresonant frequency, the net
impedance of the tuned circuit is only resistive and say it is Rar (resistance at anti resonance). The
generator sees a load of Rar at anti resonance. ~ is independent of Land C values of the tuned
circuit. This value of Rar may be greater than or less than the generator output resistance for
maximum power transfer. The resistive impedance into which the generator supplied power can be
reduced by tapping the external generator connections across only a portion of the impedance is
shown in Fig. 6.19.
x
h
L
c
R
y
(8)
(b)
Fig. 6.19 (a) Equivalent circuit with current source
Let the mutual inductance between LA and LB is such that the total iinductance is,
L=LA +LB +2M
Anti resonance will occur between points P and Y if,
..... (19)
227
Tuned Amplifiers - II
0)
(LA + La + 2M) = -
1
..... (20)
O)c
Consider the terminals P and Y of Fig. 4.35(a) :
Anti resonance will occur if
1
0)
(La + M) = -O)c
-
0)
(LA + M)
..... (21)
Because tapping is done in the inductor, part of the inductor is in series with capacit.cr C. But
still, the net reactance must be capacitive for this branch.
The circuit equations in matrix force for the Fig. 6.18 (a) are:
V] [ R B + jX LB
[ g (R B + jX LB + jX M)
= -
- (R B + jX LB + jX M )
+ jX L B + j2X M - jX C )
(R A + RB + jX LA
1[II~ ]
..... (22)
(RB + jXLB)(R A +RB + jX LA + jX LB + j2X M - jX C )
-(RB + jX LB + jxMf
..... (23)
If
..... (24)
If the Q factor of the circuit is high, and the circuit is in anti resonance, then XL = Xc.
XLB » RB. So RB can be neglected.
="X
Z
P,Y
J LB
+ (XLB +xMf
R
..... (25)
If Q value is reasonable, XLB is small compared to (X LB + XM)2 / R. So the term XLB can be
neglected. Comparing the value of impedance Z2, 3 with resonant impedance across the terminals X
and Y, where
..... (26)
228
Electronic Circuit Analysis
Then
..... (27)
Again if Q is high ( Q
= ro: ) , R must be small, or R «XL'
Then simplifying,
..... (28)
Considering Fig. 6.19 (b), if the circuit capacitance is split into two capacitors in series,
equivalent in capacitance to the single capacitor C, and if the external generator is tapped between the
two capacitors,
since
1
roL=roC'
..... (29)
and
..... (30)
The effect of tapping down in the capacitive side of the circuit is
..... (31)
This shows that impedance is reduced.
These methods become very convinient at high frequencies.
6.6.3
Reactance L Section for impedance transformation :
R is the load resistance Rio"is the net input resistance for the generator. R < Rio' R is to be matched
with Rio through impedance transformation. So two reactances of opposite sign i.e., XL and Xc are
to be connected as shown in Fig. 6.20. Land C elements are connected in the shape of L. So it is
called as L-section circuit. At a particular frequency, the oppposite reactances XL and Xc and R can
be transformed to match Rio'
Tuned Amplifiers - II
229
r-------p------~
R __- - .
L
r::
R
Ie
s
'-----Q
Fig. 6.20 Reactive L section for impedance transformation (R < Ra)
The RLC section of the circuit is a parallel resonant circuit. At anti resonance, it appears as a
resistance load for the generator. The value of the resistance load depends upon LlC ratio such that
Rin is matching with Ra for the parallel RLC resonant circuit,
ro=
R0
R2
--LC L2
Resistance at anti resonance,
..... (32)
~ = R;n = ~
CR
L=RinRC
Substituing this value ofL in eqn. (ii),
..... (33)
R. -R
10
roC =
Value of capacitance C needed for the L-section is,
l~1
C = ro Rin
fIt-
Similarly from eqn. (ii),
L
C=-R. R
10
Substituting this value in eqn. (32) for ro2,
Electronic Circuit Analysis
230
~R.10 R-R 2
ooL=
.
L= -R~'
-1!l.-1
..... (34)
R
00
So L is the value of inductance needed for the L section to ensure the desired value of load
where R < ~n'
6.6.4
Rm
Image Impedances : Reactance Matching :
The impedances ZI' Z2 and Z3 are arranged in the form of T-Network Va is the generator having
internal impedance Zli' The load impedance is Z2i' The circuit is shown in Fig. 6.21.
B
A
z.
k
Z2
Z)
A
J
Z2i
B
Fig. 6.21 T-Network -Image Impedances
The generator supplies power into the terminals AA. So the impedance betweeen terminals AA
must be equal to the generator impedance. The impedance looking into terminals BB must be equal to
load Z2i' The impedance at AA looking into one direction is the image impedance of the impedance
looking in the other direction. Zli is called the impedance of the network.
Similarly at B, B terminals the impedance looking is one direction is the same as that looking in
the other so that Z2i is also an image impedance at B, B terminals. The network is then said to be
matched on an image basis.
The value of image impedance of the T-section is computed as shown below.
The impedance ZI in between A, A terminals is,
..... (35)
The impedance looking into B, B terminals is required to be Z2i'
Z2'
I
=
Z2 +
Z3 (ZI + Zi)
ZI +Z3 +Zli
..... (36)
231
Tuned Amplifiers - II
Solving for Zli and Z2i
..... (37)
7_. =
"21
..... (38)
If impedance is measured between terminals AA of the T-section, and terminals B, B are open
circuited,
..... (39)
Similarly if impedance is measured at A, A terminals with B, B terminals short circuited,
Zl + Z2 + Z2 Z3 + Z3 Zl
..... (40)
Z2 +Z3
So the image impedance ZI i =
~ZlOC ZlSC
Similarly for the measurements made at B, B terminals gives
Zzc
=
~Z20C Z2SC
So a porperly designed T network will have the property of transformation of an impedance to
produce matching of a load and a source.
6.6.5
Reactance T Networks for impedance transformation :
In the T-network shown in Fig. 6.22 the T Network elements are reactances only, either capacitive or
inductive. Va is a generator with internal resistance R I. It is connected to a lod R2 through
T-Network.
R _ _-,
A
~
__ Q_____
~
J
________ S
Fig. 6.22 Reactance T-Network
Electronic Circuit Analysis
232
If the generator is to transfer maximum power to the laod, it is necessary only that the image
impedance Zli at terminals P,Q must be equal to R I. In other words, the load impedance R2 is
transformed by the T-network to a value at the P,Q terminals equal to R I.
When R2 is connected, the image impedance Zli must be
..... (41)
X may have either positive sign or negative sign, i.e., it may be either inductive or capacitive in
nature. Then, RI R2 + j RI (X2 + X3)
= -
(Xl X2 + X2 X3 + X3 Xl) + jR2 (Xl + X3)
By equating real terms,
..... (42)
(Xl X2 + X2 X3 X3 + Xl)
In the above equation, RI R2 term on the LHS is positive. So the term on RHS must also be
positive. Since it is negative sign, one or more of the three terms on RHS must be positive. X can be
RI R2
+jWL or
;"b .
= -
In order that the product term Xl X2 is positive, Xl can be +jXa and X2 can be -jXa
:-jXb) = +Xa Xb· This requires that one reactive arm of the T-Network be opposite in sign to the sign
)f the other two arms. So the T-Network must consist of one capacitance and two inductances or
Vice-versa.
By equating imaginary terms,
RI (X2 + X3) = R2 (Xl + X3)
..... (43)
The above equation may be written as,
RI R2 = - [(Xl + X3) (X2 + X3) - Xl]
..... (44)
..... (45)
Tuned Amplifiers - II
233
So the value of one of the reactance arms is,
-Xl
=
-X3
±
Rl ( 2
R2 X3 -R 1 R2
)
..... (46)
Substituting eqn. (46) in (44),
or
..... (47)
The above equations are the design equations for the T-Network, in terms of the values ofX 3.
234
Electronic Circuit Analysis
1. The purpose of resonant circuits in tuned circuits is _ _ _ _ _ _ _ __
2. Give the classification of tuned amplifiers.
3. What are the applications of tuned amplifiers?
4. Give the frequency response of a tuned amplifier.
5. What is the expression for Rp in terms of ro, L, R ? What is Rp ?
6.
What is a called? What is the expression in terms of ro, roo ?
7. What isAIAo?
8. Why do we go for tapped single tuned amplifier?
9: What is maxiuni power transfer theorem ?
10. What is a double tuned amplifier?
II. What is staggered tuning ?
12. At what frequencies are the tuned amplifiers operate?
13. What is a tank circuit ?
14. What is the expression for harmonic distortion in tuned amplifiers ?
15. How do you build a class B tuned amplifier?
Tuned Amplifiers - II
1.
Draw the circuit for BJT tuned class B/C amplifier. Explain its working.
2.
Draw the circuit for JFET tuned R.F. amplifier and explain its working.
3.
Explain the principle and working of wide band amplifiers.
235
Electronic Circuit Analysis
236
1.
(a) To provide properly matching load impedance
(b) To reset unwanted harmonics
(c) To couple power to load.
2.
Single tuned, double tuned, stagger tuned amplifiers.
3.
RF amplifiers, communication receivers.
4.
5.
.1\,
e I\, is the series internal resistance of inductor represented as a shunt
R'
ro 2
=
element.
-.
6.
o- fractional frequency variation. 0 =
7.
A
1
=---Ao 1+ j2<>Q e
8.
9.
ro - roo
roo
--~
10.
For impedance matching.
For a linear circuit, for the maximum power to he delivered to the load the output or load
impedance should be the complex conjugate of the effective or equivalent input impedance
of the circuit.
It has two resonant circuits both tuned to the same frequency.
11.
Two resonant circuits tuned to different frequencies.
12.
13.
14.
High or radio frequencies.
Parallel LC circuit is called a tank circuit.
There is very negligible harmonic qistortion in tuned amplifiers as they are narrow band
amplifiers.
15.
By removing the bias resistors from class A amplifier, we can build a class ~ amplifier.
UNIT - 7
Voltage Regulators
In this Unit,
• Different types of Voltage Regulator Circuits are explained.
• Terminology associated with Voltage Regulator Circuits is given.
• I.C. Voltage Regulator Circuits, 3 terminal Voltage Regulator Circuits
are explained.
• Voltage Multiplier Circuits, Voltage Doubler Circuits are also given.
• Voltage TripIer, quadrupler circuits are explained. Numerical examples
are also given.
238
7.1
Electronic Circuit Analysis
Introduction
Voltage Regulator Circuits are electronic circuits which give constant DC output voltage, irrespective
of variations in Input Voltage Vi' current drawn by the load IL from output terminals, and
Temperature T. Voltage Regulator circuits are available in discrete form using BJTs, Diodes etc and in
IC (Integrated Circuit) form. The term voltage regulator is used when the output delivered is DC
voltage. The input can be DC which is not constant and fluctuating. If the input is AC, it is converted
to DC by Rectifier and Filter Circuits and given to I.C. Voltage Regulator circuit, to get constant DC
output voltage. If the input is A.C 230 V from mains, and the output desired is constant DC, a
stepdown transformer is used and then Rectifier and filter circuits are used, before the electronic
regulator circuit. The block diagram~ are shown in Fig. 7.1 and 7.2.
+
Unregulated
A.C. Input
Step Down
Transformer
Rectifier
Circuit
Filter
Circuit
Electronic
Regulator
Circuit
1--.,.-- Regulated
RL
D.C.
Output
Voltage
Fig. 7.1 Block Diagram of Voltage Regulator with AC Input
--~---{----;E~le-c~tro~n~ic----l-------------rAr+
Un regulated
D.C.lnput
Regulator
Circuit
V.1
Fig. 7.2 Block Diagram of Voltage Regulator with D.C Input
The term Voltage Stabilizer is used, if the output voltage is AC and not DC. The circuits used
for voltage stabilizers are different. The voltage regulator circuits are available in IC form also. Some
of the commonly used ICs are, J.1A. 723, LM 309, LM 105, CA 3085 A.
7805, 7806, 7808, 7812, 7815 : Three terminal positive Voltage Regulators.
7905, 7906, 7908, 7912, 7915 : Three terminal negative Voltage Regulators.
The Voltage Regulator Circuits are used for electronic systems, electronic circuits, IC
circuits, etc.
The specifications and Ideal Values of Voltage Regulators are :
Specifications
Ideal Values
1. Regulation (Sv)
2. Input Resistance (R.)
1
3. Output Resistance (Ro)
4. Temperature Coefficient CST)
5. Output Voltage V0
6. Output current range (lL)
7. Ripple Rejection
0%
<Xl ohms
o ohms
o mv/oc.
0%
Voltage Regulators
7.1.1
239
Different types of Voltage Regulators are
1. Zener regulator
2. Shunt regulator
3. Series regulator
4. Negative voltage regulator
5. Voltage regulator with foldback current limiting
6. Switching regulators
7. High Current regulator
7.1.2
Zener Voltage Regulator Circuit
A simple circuit without using any transistor is with a zener diode Voltage Regulator Circuit.
In the reverse characteristic voltage remains constant irrespective of the current that is flowing
through Zener diode. The voltage in the break down region remains constant. (Fig. 7.3)
Fig. 7.3 Zener diode reverse characteristic
Therefore in this region the zener diode can be used as a voltage regulator. If the output voltage
is taken across the zener, even if the input voltage increases, the output voltage remains constant. The
circuit as shown in Fig. 7.4.
The input Vi is DC. Zener diode is reverse biased.
+
V
I
Zener
Diode
Fig. 7.4 Zener regulator circuit
If the input voltage Vi increases, the current through Rs incre~ses. This extra current flows,
through the zener diode and not through RL. Therefore zener diode resistance is much smaller than
RL when it is conducting. Therefore IL remains constant and so Vo remains constant.
Electronic Circuit Analysis
240
The limitations of tlris circuits are
I. The output voltage remains constant only when the input vo\t€lge is sufficiently large so that
the voltage across the zener is Vz.
2. There is limit to the maximum current that we can pass through the zener. If Vi is increased
enormously, I z increases and hence breakdown will occur.
3. Voltage regulation is maintained only between these limits, the minimum current and the
maximum permissible current through the zener diode. Typical values are from 10m A
to I ampere.
7.1.3
Shunt Regulator
The shunt regulator uses a transistor to amplify the zener diode current and thus extending the
Zener's current range by a factor equal to transistor hFE . (Fig. 7.5)
+,
Unregulated
D.C.
RI
YEB
Is
Rs
II
V.1
A
Iz
II
10
Q
r
Yo
IB
Fig. 7.5 Shunt Regulator Circuit
Zener current, passes through R)
Nominal output voltage
= V z + V EB
The current that gets branched as IB is amplified by the transistor. Therefore the total current
10 = (~ + I) IB, flows through the load resistance RL. Therefore for a small current through the
zener, large current flows through RL and voltage remains constant. In otherwords, for large current
through Ru Vo remains constant. Voltage Vo does not change with current.
Example : 7.1
For the shunt regulator shown, determine
1.
2.
3.
4.
5.
The nominal voltage
Value ofR),
Load current range
Maximum transistor powt::r dissipation.
The value ofRs and its power dissipation.
Vi
=
Constant. Zener diode 6.3V, 200mW, requires SmA minimum current.
Transistor Specifications:
VEB
=
O.2V, hFE
=
49, leBo
=
O.
Voltage Regulators
241
I. The nominal output voltage is the sum of the transistor VEB and zener voltage.
V0 = 0.2 + 6.3 = 6.SV = VEb + Vz
2. RI must supply SmA to the zener diode:
R
I
=
8V -6.3
SxlO-3
1.7
Sx10-3
=
340 0
3. The maximum allowable zener current is
Power rating
0.2
- - - - .=-=- = 31.8 mA
Voltage ratmg 6.3
The load current range is the difference between minimum and maximum current through the
shunt path provided by the transistor. At junction A, we can write,
IB = Iz - II
II is constant at 5 rnA
Is = Iz - II
I I is constant at 5 rnA
Is = 5 x 10-3 - 5 x 10-3 = 0
Is = IZ Max - 12
= 31.8 x 10-3
-
5
x
10-3 = 26.8 rnA
The transistor emitter current IE = Is + Ie
Ie = ~ Is = hFE Is
m
IE =
+ I) Is = (hFE + I) Is
Is ranges from a minimum of 0 to maximum of 26.8 rnA
Total load current range is (hFE + I) Is
= 50 (26.8 x 10-3) = 1.34 A
4. The maximum transistor power dissipation occurs when the current is maximum IE ::::: Ie
Po= Vo IE = 6.5 (1.34) = 8.7 W
5.
Rs must pass 1.4 A to supply current to the transistor and RL .
8-6.5
=1.120
1.34
The power dissipated by Rs,
Rs
= (1.34)2 . (1.12) = 24 W
= IS2
Regulated Power Supply
An unregulated power supply consists of a transformer, a rectifier, and a filter. For such a circuit
regulation will be very poor i.e. as the load varies (load means load current) [No load means no load
Electronic Circuit Analysis
242
current or 0 current. Full load means full load current or short circuit], we want the output voltage
to remain constant. But this will not be so for unregulated power supply. The short comings of the
circuits are :
1. Poor regulation
2. DC output voltage varies directly as the a.c. input voltage varies
3. In simple rectifiers and filter circuits, the d.c. output voltage varies with temperature also,
if semiconductors devices are used.
An electronic feedback control circuit is used in conjuction with an unregulated power supply
to overcome the above three short comings. Such a system is called a "regulated
power supply".
Stabilization
The output voltage depends upon the following factors in a power supply.
1. Input voltage Vi
2. Load current IL
3. Temperature
Change in the output voltage ~Vo can be expressed as
avo
avo
avo
I
aiL
aT
+ Ro
~IL
~V = --.~V.+--.~IL +--.~T
o
aVi
~Vo = SI ~Vi
+ ST
~T
Where the three coefficients are defined as,
(i)
Stability factors.
This should be as small as possible. Ideally 0 since V0 should not change even if Vi changes.
(ii)
(iii)
Output Resistance
-
Temperature Coefficient
The smaller the values of the three coefficients, the better the circuit.
243
Voltage Regulators
7.1.4
Series Voltage Regulator
The voltage regulation (Le., change in the output voltage as load voltage varies (or input voltage
varies) can be improved, if a large part of the increase in input voltage appears across the
control transistor, so that output voltage tries to remain constant, i.e., increase in Vi results in increased
VCE so that output almost remains constant. But when the input increases, there may be some
increase in the output but to a very smaller ~xtent. This increase in output acts to bias the control
transistor. This additional bias causes an increase in collector to emitter voltage which will compensate
for the increased input.
If the change in output were amplified before being applied to the control transistor, better
stabilization would result.
Series Voltage Regulator Circuit is as shown in Fig. 7.6 .
.. Series Voltage Regulator Circuit
r
V.1
-
1
~
1
dY.=V
'
,
-
________-L________
~
__
~~~
__
~~~
__
~
____
~
__
~
Fig. 7.6 Series Regulator Circu.it
Q 1 is the series pass element of the series regulator. Q2 acts as the difference amplifier. D is the
reference zener diode. A fraction of the output voltage b Vo (b is a fraction, which is taken across R2
and the potentiometer) is compared with the reference voltage YR' The difference (bVo - VR) is
amplified by the transistor Q2' Because the emitter of Q2 is not at ground potential, there is constant
voltage YR' Therefore the net voltage to the Base - Emitter of the transistor Q2 is (bVo - VR). As Vo
increases, (bVo - VR) increases. When input voltage increases by !::.Vi' the base-emitter voltage ofQ2
increases. So Collector current of Q2 increases and hence there will be large current change in R3,
Thus all the change in Vi will appear across R3 itself. VBE of the transistor Q 1 is small. Therefore the
drop across R3 = VCB ofQ 1 ::::: VCE ofQ 1 since VBE is small. Hence the increase in the voltage appears
essentially across Q 1 only. This type of circuit takes care of the increase in input voltages only. If the
input decreases, output will also decrease. (If the output were to remain constant at a specified value,
even when Vi decreases, buck and boost should be there. The tapping of a transformer should be
changed by a relay when VI changes). ro is the output resistance of the unregulated power supply
which proceeds the regulator circuit. ro is the output resistance of the rectifier, filter circuit or it can
be taken as the resistance of the DC supply in the lab experiment.
Electronic Circuit Analysis
244
flV
The expression for Sv (Stability factor) = fl V~
1
=
~=
[Rl + R 2 ]. (Rlin parallel with R 2 )+ h ie2 +(1 + h fe2 )R z
R2
hfu2 R 3
Zener diode resistance (typical value)
rO +
R3 + h ie1
1+ h fel
The preset pot or trim pot R3 in the circuit is called as sampling resistor since it controls or
samples the amount of feedback.
Preregulator
It provides constant current to the collector of the DC amplifier and the base of control element.
If R3 is increased, the quantity
R3 + h ie1
also increases, but then T is very small, since it is
1+ hfe
being divided by hfe .
7.1.5
Negative Voltage regulator
Sometimes it is required to have negative voltage viz., --6V, -ISV, -21 V etc with positive terminal
grounded. This type of circuit supplies regulated negative voltages. The input should also be
negative DC voltage.
7.1.6
Voltage regulator with fold back current limiting
In high current voltage regulator circuits, constant load current limiting is employed. i.e. The load
current will not increase beyond the set value. But this will not ensure good protection. So foldback
current limiting is employed. When the output is shared, the current will be varying. The series pass
transistors will not be able to dissipate this much power, with the result that it may be damaged. So
when the output is shorted or when the load current exceeds the set value, the current through the
series transistors decrease or folds back.
7.1.7
Switching regulators
In the voltage regulator circuit, suppose the output voltage should remain constants at +6V, the input
can be upto + 12V or + 15V maximum. If the input voltage is much higher, the power dissipation
across the transistor will be large and so it may be damaged. So to prevent this, the input is limited to
around twice that of the input. But if, higher input fluctuations were to be tolerated, the voltage
regulators IC is used as a switch between the input and the output. The input voltage is not connected
permanently to the regulator circuit but ON/OFF will occur at a high frequency (50 KHz) so that
output is constantly present.
A simple voltage regulator circuit using CA3085A is as shown in Fig. 7.7. It gives 6 V constant
output upto 100 rnA.
Voltage Regulators
245
+ '13Y .....,~-.....--~
2
3
Un regulated
input
Fig. 7.7 CA 3085A IC Voltage Regulator
~Av
Vo=
~
=
~+~Av)
R2
Rl+R2
RCA (Radio Corporation of America) uses for the ICs, alphabets CA. CA3085A is voltage
regulator. LM309 K - is another Voltage Regulator IC.
!lA716C is head phone amplifier, delivers 50 mW to, 500-6000 load
CA3007 is low power class AB amplifier, and delivers 30 mW of output power.
MC1554 is 20W class B power amplifier.
Preregulator
The value of the stability factor Sy ofa voltage regulator should be very small. Sy can be improved
tV. - VO)
if R3 is increased (from the general expression) 'since R3 :::::
1
• We can increase R3 by
I
decreasing I, through R3. The current I through R3 can be decreased by using a Darlington pair for
Q!. To get even better values of Sv' R3 is replaced by a constant current source circuit, so that
R3 tends to infinity (R3 ~ 00). This constant current source circuit is often called a transistor
preregulator. Vi is the maximum value of input that can be given.
Short Circuit Overload Protection
Overload means overload current (or short circuit). A power supply must be protected further from
damage through overload. In a simple circuit, protection is provided by using a fuse, so that when
current excess of the rated values flows, the fuse wire will blow off, thus protecting the components.
This fuse wire is provided before roo Another method of protecting the circuit is by using diodes.
IL
+
From amplifier Q of series voltage regulator
2
Fig. 7.8 Circuit for short circuit protection
Electronic Circuit Analysis
246
Zener diodes can also be employed, but such a circuit is relatively costly.
The diodes D] and D2 will start conducting only when the voltage drop across Rs exceeds the
current in voltage of both the diodes D] and D2. In the case of a short circuit the current Is will
increase upto a limiting point determined by
V
Is
Y1
=
+ V - VBE1
Y2
R
S
When the output is short circuited, the collector current of Q2 will be very high Is . Rs will
also be large.
The two diodes D] and D2 start conducting.
The large collector current of Q2 passes through the diodes D] and D2 and not through
the transistor Q].
Transistor Q] will be safe, D] and D2 will be generally si diodes, since cut in voltage is
0.6V. So Is Rs drop can be large.
- Example 7.2
Design a series regulated power supply to provide a nominal output voltage of 2SV and supply load
current IL ~ lA. The unregulated power supply has the following specifications Vi = SO ± SV,
and ro = 10 n.
Given: Rz = 12 nat Iz = 10 rnA. At IC2 = 10 rnA,
p = 220, hie2 = 800 n, hfe2 = 200, I]
The reference diodes are chosen such that VR:::
-f
= 10 rnA.
Vo
"2'
V
=
12.5 V
.. Two Zener diodes with breakdown voltages of7.S V in series may be connected.
Rz = 12 n at Iz = 20 rnA
Choose
IC2 ::: IE2 = 10 rnA
At Ic = 10 rnA, the h-pararneters for the transistor are measured as,
2
p = 220, hie2 = 800 n, hfe2 = 200
Choose
ID = lOrnA, so that Iz = ID] + ID2 = 20 rnA
R =
Vo - VR
ID
IB2
2
-p= 220
D
IC
=
25-15
10
=IKn
lOrnA
=
4SIlA
Choose I] as 10 rnA for si transistors, VBE = 0.6 V
V2 = VBE2 '+ VR = 15.6 V
R
]
=
Vo - V2
II
=
25-15.6
IOxlO- 3
= 940 n
Voltage Regulators
247
R
V
- 2
2- II
~
=
15.6
lOx 10-3
= 1,560 Q
For the transistor QI' choose 12 as IA and hFEI
Is I
=
=
125 (d.c. current gain ~)
IL +11 +ID
h fn)
fel \P
(DC Current gain)
1000+ 10+ 10
125
::: 8 rnA
The current through resistor R3 is I = lSI + IC2 = 8 + 10 = 18 rnA
The value of R3 corresponding to Vi = 45 V and IL = IA is (since these are given in the
problem Vi = 50 + 5, 45 V)
Y.1 -(VBE I +
I
Yo)
50-25.6
3 = 1,360 n.
18x10Voltage regulator is a circuit which maintains constant output voltage, irrespective of the changes
of the input voltage or the current.
Stabilizer - If the input is a.c, and output is also a.c, it is a stabilizer circuit.
R3 =
7.2
=
Terminology
Load Regulation: It is defined as the % change in regulated output voltage for a change in load
current from minimum to the maximum value.
EI = Output voltage when IL is minimum (rated value)
E2 = Output voltage when IL is maximum (rated value)
% load regulation
EI -E 2
EI
x 100 %; EI > E2. This value should be small.
Line Regulation : It is the % change in Vo for a change in VI'
~Vo
= - - x 100 %
~VI
In the Ideal case
~V 0 =
0 when V0 remains constant.
This value should be minimum.
Load regulation is with respect to change in IL
Line regulation is with respect to change in Vi'
Ripple Rejection
ripple voltage.
It is the ratio of peak to peak output ripple voltage to the peak to peak input
V~ (p - p) Vo' = output ripple voltage
V; (p - p)
Vi' = input ripple voltage
Electronic Circuit Analysis
248
Stand by Current Drain
It is the current drawn by the regulator circuit, when IL = 0
(current drawn by the circuit components only and not by the load)
Short Circuit Current Limit :
The output current of the regulator (IL) when the output terminals are shorted.
Sense Voltage :
It is the voltage between current sense and current limit terminals.
Temperature Stability or Average Temperature Coefficient :
Change in Vo per unit change in temperature m Vloc'
7.3
Basic'Regulator Circuit
A monolothic voltage regulator circuit mainly consists of 3 parts.
1. Reference voltage circuit
2. Error Amplifier
3. Series pass element
~AV
V = -----''-o 1+ ~AV
V.
1
---------1 Series Pass element 1-------.----,..
Fig. 7.9 Basic I.C. voltage regulator circuit
Voltage reference circuit generates a constant voltage level. A fraction of the output voltage is
derived by the potential divider network R 1, R2 and this voltage is compared with the reference
voltage. The difference in these two voltages is converted into an error signal, which will control the
voltage drop across the series pass element, to keep V0 constant..
If Vo is less, the drop across series pass element is reduced, so that Vo increases. If Vo
is more, the drop across series pass element increases so that V 0 decreases and comes to the
normal value.
Voltage Regulators
====
The error amplifiers controls the base current to the series pass element, which is a transistor
and the drop across it, VeE varies, as its Ie changes, in proportion to lB. The characteristics of a
good regulator circuit are :
1. It should have low line regulation
2. It should have low load regulation
3. It must have a high degree of ripple rejection
Voltage regulator ICs are provided with 3 kinds of protection
1. Short circuit protection
(a) Active current limiting:
(b) Passive current limiting: This isjoldback current limiting
2. Over voltage protection
3. Thermal overload protection
7.4
Short Circuit Protection
Vc
IB2
\1
B
Current
limit terminal
Vo
-
X
V
z
Current
sense terminal
Fig. 7.10 Short circuit protection circuit
7.4.1 Active Current Limiting
Normally T j is off. A resistor Rse is connected between current limit and current sense terminals.
When Vo is shorted, the drop across Rse is such that T J turns ON. When T J turns ON, it draws
current (Ie flows) and reduces the base drive IB to the power device T 2 .
:. 10 is reduced to zero.
If the short circuit current limit is 65 rnA, and cut in voltage VBE for T) is 0.65 V, Rse must
be suc~ that when 65 rnA is reached T) should tum ON.
0.65V
Rse = 65mA = IOn
Current sense will detect the short circuit. Short circuit current will flow through this terminal.
7.4.2
Foldback Limiting
If the short circuit is not detected, and short circuit condition exists for a long time, due to large
current flow, excessive heating will take place, damaging the I-C. In such cases, foldback limiting is
employed. When short circuit occurs, IL decreases and becomes a minimum value.
Electronic Circuit Analysis
250
The circuit is as shown:
Regulated O/P
Unregulated
input
CL current limit
CS current sense
T)
Fig. 7.11 FoJdback limiting circuit
T/ and T2 are normally off. When IL increases to a high value, the voltage drop across
Rse increases and turns on T J • T J inturn, turns on T2. Since Ie ofT J = IB ofT2, when T2 turns ON,
the base drive for T3 is reduced. Therefore Ie of T3 i.e., IL reduces.
7.4.3 Over Voltage Protection
Here Zener diode is connected between output and ground terminals. When Yo decreases, the diode
starts conducting, maintaining Yo constant at the breakdown voltage of the zener.
Voltage Regulators
251
1. The stability factor Sy in Voltage Regulator Circuits is defined as _ _ _ _ _ _ __
2.
Output Resistance Ro of Voltage Regulator Circuit is defined as _ _ _ _ _ _ __
3.
Temperature coefficient of Voltage Regulator Circuits is defined as _ _ _ _ _ _ __
4.
The present pot or trimpot element in the series Voltage Regulator Circuit is called as
5.
The purpose of Pre regulator circuit is the series Voltage Regulator Circuit is _ _ _ _ _ __
6.
Fold back limiting circuit is also known as _ _ _ _ _ _ __
7.
In a voltage stabilizer circuit, the input and outputs are _ _ _ _ _ _ __
8.
Load Regulation is defined as _ _ _ _ _ _ __
9.
Line Regulation is defined as _ _ _ _ _ _ __
10. Ripple Rejection is defined as _ _ _ _ _ _ __
11. The three important sections of Voltage Regulator LC.s are _ _ _ _ _ _ __
12. The elements used in Active Current limiting circuit of Voltage Regulators are
Electronic Circuit Analysis
252
====================::jl
.F====================
1. Draw the circuit for shunt type Voltage Regulator and explain its working.
2. Explain the terms
(i)
(ii)
(iii)
(iv)
Stabilization
Stability factor Sy
Output Resistance Ro
Temperature Coefficient, pertaining to Voltage Regulators.
3. Draw the circuit for series type voltage regulator and explain its working.
4. What is the function of pre regulator circuit? Explain.
5. Draw the circuit and explain how short circuit over load protection is provided in Voltage
Regulator circuits.
6. Define the terms
(i)
(ii)
(iii)
(iv)
(v)
Load Regulation
Line Regulation
Ripple Rejection
Sense Voltage
Temperature stability pertaining to Voltage Regulator ICs.
Voltage Regulators
253
4.
Sampling Resistor
5.
It provides constant current to the collector of the d.c. amplifier and base of control element.
6.
Crow - bar limiting circuit
7. a.c.
8.
(VI - V 2N I)
9.
I1V
( 11VI
o)
X
x
100 %
100 %
VJ(p-p)
10.
vi(P-p)
11. (a) Reference Voltage Circuit
(b) Error Amplifier
(c) Series Pass Element
12. Transistors, Resistors, Zener Diodes.
"This page is Intentionally Left Blank"
UNIT - 8
Switching and
Ie Voltage Regulators
In this Unit,
• IC 723 Voltage Regulators and 3 terminal IC Regulators
• Current Limiting
• Specifications of Voltage Regulator Circuits
• DC to DC Converters
• Switching Regulators
• Voltage Multipliers
• UPS and SMPS
256
8.1
Electronic Circuit Analysis
IC 723 Voltage Regulators and 3 Terminal IC Regulators
1. Current sense
2. Inverse input
3. Non Inverse input
4. Vref
5. -Vee
6. YOU!
7. Ve
8. + Vee
9. Frequency compensation
10.Current limit
V- (pin 5) can be grounded 8 and 7 are shorted and DC Viis given.
VR = V23 = 5 V
R = 10 n (Ohms)
IL :::: 0.5 A
Frequency
compensation
Fig. 8.1 Ie 723 Voltage Regulator
8.1.1
IC 723 Pin Configuration
The pin configuration of the 723 I.C. voltage regulator is shown in Fig. 8.3
Pin 1: Current sense. The drop across Rse connected between this pin and pin 10 acts as VBE of
the current limit transistor.
Pin 2: Inv. input. It is the inverting input terminal of the error amplifier of the I.C., in the
schematric circuit.
Pin 3 : Non-Inv. input:
it is the non-inverting terminal of the error amplifier.
Pin 4: Vref: It is the reference voltage terminal of the reference source provided in the I.C.
Pin 5 : -V: Negative Voltage is applied at this terminal.
Pin 6: Regulated output voltage V0 is obtained at this terminal.
Pin 7: Control Voltage Terminal.
Pin 8 : Positive Voltage is applied to this terminal.
Pin 9: Frequency Compensating Capacitor is Connected at this pin.
Pin 10: Short circuit current limiting resistor Rse is connected between pin nos 10 and 1.
257
Switching and Ie Voltage Regulators
723 I.C. Voltage Regulator can give a load current of ISO rnA on its own. By using external pass
transistor the IL can be enhanced to several amperes.
The internal reference voltage is approximately 7.ISv. Resistors R) and ~ are used to set the gain
of the internal amplifier.
Resistors R3 acts as DC bias compensation provider for R) and R2 •
If the output voltage desired is less than the reference voltage Vref ' (R) potential divider, and the desired Vo «V re f) can be
. obtained.
~)
network acts as
The current limit is set by,
I limit =
V
sense
Rsc
Vsense is approximately 0.6SV at room temperature. Rsc resistor is placed in series with load and
-. so the current through it is the load current.
The current limit transistor is used such that the drop across Rsc is applied as VBE for the transistor.
The collector of the current limit transistor is conected to the base of the output pass transistor. If
the output current rises to the point where the voltage across Rsc exceeds 0.6SV, the current limit
transistor wil turn ON, so the output current is shunted away from the output pass transistor.
723 I.C. is available in (1) Metal can package and (2) Duel-in-line packages (DIP). The pin
configurations are shown in Fig. 8.2 and 8.3.
I. Current sense
2. Inverting input
3. Non-Investing input
4. VRef
S. V-
6. Vo
7.Vc = Control Voltage
Top view
8. V+
9. Frequency Compensation
10. Current limit.
Fig. 8.2 723 Ie Metal can Package
258
Electronic Circuit Analysis
No Connection (NC) 1
14 NC
Current Limit 2
13 Frequency compesation
Current Sense 3
12 y-
723
Inverting Input 4
Non-Inverting Input 5
YRef 6
Y- 7
8 Nc
~------------------~
Top view
Fig. 8.3 723 IC DIP Package
The internal schemetric of the 723
Ie is shown in Fig.
8.4
Y+
Frequency
compensation
7
9
Yc
8
Temperature
Compesated
Zener
>----0
4
Yrcf
5
Current
Limiting
Transistor
Y-
Current
Limit
10
Current
Sense
Fig. 8.4 Internal schematic of 723 voltage regulator I.C.
259
Switching and Ie Voltage Regulators
Problem 8.1 : Design a + 12 V voltage regulator circuit using 723 Ie to give current limit of 50 mAo
Vo=Vref.(Rl:2R2]
Let R2 =12 KO . Vref = 7.15v. Vo =12v, RI =?
Vo
RI=-·R2-~
Vref
12
= -.12-12
7.15
= 20.14 - 12 = 8.14::: 8KO
12 x8 96
R3 = Rill R2 = 12+8 = 20 = 4.8KO
The circuit diagram is shown in Fig. 8.5
vm
V+
6
12
11
Vo
Rsc
CL
723
R=4kn
CS
14
100 pF
Fig. 8.5 Circuit diagram for a +12V voltage regulator circuit.
8.1.2
3 - Pin Voltage Regulator les
7800 I.e series is of3- pin positive voltage regulator les 7900 Ie series is of3 - Pin Negative Voltage
regulator les.
3 - Terminal Voltage Regulators
7800 Series Voltages Regulators:
Fig. 8.6 Three terminall.C. Regulator Circuit
Electronic Circuit Analysis
260
These are 3 terminal, positive voltage regulators. Vo is positive. General circuit is :
XX: indicates the numbers that will follow like 00, 01, 02 etc.,
These are available in TO - 3 type Metal package.
Pin 1
Input
2: Output
Pins 1 and 2 can be
Case : Ground
known from the base diagram
TO 220 type Plastic package
Pin I
Input
2
Ground
3
Output
Fig. 8.7 Pin diagram
Device type
Maximum input voltage (V)
Output voltage V0
7805
5.0
7806
. 6.0
7808
8.0
78012
12.0
7815
15.0
7818
18.0
7824
24.0
35V
40V
I
7805 can be used as a 0.5 A current source. The current supplied to the load is,
V
IL
IQ
=
RR
+ IQ
= Quiescent current, IL = Load current = 4.3 rnA for 7805
7800 series positive voltage regulator ICs.
Ie No.
7802
7805
7808
7818
7824
Vo
+2V
+5V
+8V
+ 18V
+24 V
Switching and Ie Voltage Regulators
261
7900 series is negative voltage regulator ICs.
ICNo.
Vo
-2V,
-5
-6
>
-8
-18
-24 I
7902
7905
7906
7908
7918
7924
MaxV j
,
-35
-40
output
Bottom view
Fig. 8.8 Pin configuration base diagram
Vin ---.-J....-.r-:;7rR:SOo.S;1--.....,...,..--=--....--..,. Output
T
Cl 0.33 IlF 2
1....-----+--:--+---'!I-~
1
Q
lOW
Fig. 8.9 Ie 7805 Voltage Regulator
VO=VR+VL
Vo = IL RL
RL = 100 (Ohms)
VL = 5 V ;
VO=VR+VL
= 5V + 5V
IL
= 0.5 A
=
lOY
v0
Electronic Circuit Analysis
262
The voltage drop across 7805 is 2V
.. The minimum input voltage required is Yin = Vo + Dropout voltage
Vin=12V
So a current source circuit using a voltage regulator can be designed for a desired value of IL'
by choosing an appropriate value of R.
Example : 8.2
Using 7805C, voltage regulator, design a current source that will
480 lOW load.
deliv~r
0.25 A current to the
Neglecting
VO=VR+VL
= 5V + (48 0) (0.25) = 17V
Vin = V0 + drop across Ie
Vin = 17 + 2 = 19V Ans.
8.2
Current Limiting
This means short circuit protection i.e. Even if the output terminals of the voltage regulator are
shorted, the current should be limited and should not exceed a particular value. This can be done
in two ways.
1. Simple limiting circuit
2. Foldback limiting circuit
8.2.1
Simple Limiting Circuit
As the drop across R3, VC2 B2 increases, 'V B2 E2 decreases.
!
Q2 goes off.
R4 is called current sensing resistor.
",.. If IL < 600 rnA, voltage drop across R4 is < 0.6V
The drop across R4 is VBE for transistor Q3'
(... R4 = 1 0,-1
x
600 rnA = 0.6 V).
:. Q 3 is cut off. So the regulator circuit works as a normal circuit, without any
limiting provision. When IL is between 600 and 700 rnA, the voltage across R4 is between 0.6 V and
0.7 V. Therefore Q 3 will turn on. So the collector current of Q3 will flow through R3. So the base
voltage VBE to Q2 will decrease. Therefore output voltage Vo will decrease and hence load current
will decrease.
Switching and Ie Voltage Regulators
263
Fig. 8.10 Current limiting circuit.
If, Is == Short circuit current when output terminals are shorted
Voltage across R4 is VBE == Is . R4
..
Is == VBE / R4
By choosing the value of R4 , we can change the level of current limiting.
The disadvantage with this circuit is power dissipation across transistor Q2 will be very large.
8.2.2
Fold Back Limiting
+
V.
m
Fig. 8.11 Foldback limiting circuit.
As IB of a transistor decreases, Ie decreases. Therefore 10 also decreases, instead of fully being cut
off. This is the principle offoldback limiting.
Using this circuit, we can reduce the power dissipation in the pass transistor. The load current
flows through R4 producing a voltage drop::: IL • R4. So the voltage fed to the potential divider circuit
Rs and R6 is, IL . R4 + Yo' This voltage controls Q3" The feedback fraction is,
264
Electronic Circuit Analysis
I
max
RL (Minimum)
Fig. 8.12 1- RL Characteristic
The maximum load current will be higher than short circuit current.
8.3
8.4
Specifications of Voltage Regulator Circuits
1. Regulation %
2. Input Z
3. Output Z
4. Ripple rejection
5. Current rating
6. Voltage rating
DC To DC Converter
These are used when large DC voltage is required from small DC voltage i.e. If DC 5V is available,
we can make it 15V. The DC 5V is used to drive an oscillator circuit. The AC output is amplified with
transformer. Then it is converted to DC, to get large DC output voltage.
j II
Rectifier
&
Filter circuit
+
Transformer
Fig. 8.13 DC to DC Converter
8.5
Switching Regulators
In the series regulators, the excess voltage is dropped across the series pass element. So power
dissipation is more. In order to reduce this, switching regulators are used. Here a switch connects the
input to the regulator intermittently, so that average current is passed to the load. When the switch is
closed, energy is stored in an inductor. This energy is transferred to the load. When the load voltage
decreases, this is sensed by the comparator and when the energy in the storage element is dissipated,
the switch is closed by the comparator and input is connected to the regulator circuit. So the storage
device gets charged again.
Switch is a transistor which is turned ON and OFF by the voltage of the threshold detector.
When switch is open, Diode D conducts. The energy stored in 'L' forward biases the
diode. This maintain current flow through. When'S' is closed, Diode D is Reverse Biased. Capacitor
C charges. It supplies energy to the load. When V0 decreases, detector changes state, and closes
the switch.
Switching and Ie Voltage Regulators
265
L
v
io
0'-----
C
TD
Threshold
detector
Fig. 8.14 Regulator switch.
Switching Regulators
Voltage Regulators
I
Non Dissipative
Dissipative
ex : Series - voltage regulator.
ex: Switching regulator.
Shunt - voltage regulator.
Dissipative - Excess voltage when Vi' increases is dissipated across a series pass element.
Efficiency (TJ) is less.
Non,.cJissipative - Switching type. Input is not permanently connected. (Efficiency) TJ is more.
Block Schematic of Switching Regulator
~)I
II---)~I
VI _ _ Rectifier
(AC)
.
.
.
Filter
I
Switching
controller
I to 100 KV
High
frequency
filter &
rectifier
)
Regulator
DC
Voltage
Fig. 8.15 Switching Regulator
The voltage regulating transistor is operated in cut off or saturation. So current flowing
through this is small. Hence power dissipation in the circuit is less. Efficiency 11 is more.
The chopped AC voltage is filtered by high frequency filter and rectified to get DC.
Switching regulator is (advantages) :
1. Lighter
2. Smaller
3. More efficient than a series-pass type.
266
8.5.1
Electronic Circuit Analysis
Step - Down (Buck) Switching Regulator
Control voltage (to tum ON/OFF the transistor Ql)
~I.
10
V
m
Fig. 8.16 Bucking type regulator
tON
Vo=
T
tON
Time period for which Q l is ON
=
Vin
Circuit when SI closed
Sl
++~~'d
--IL
n!v.
Circuit when SI open
Fig. 8.17 Circuits and waveforms
T
=
Total time period Ql of input square wave.
D = Duty cycle
tON
---=~-
tON + toff
tON
tON
Total time
T
= --='-'--- = - -
D1 is catch diode in the circuit
It provides continuous path for the inductor current when Q l turns off. When SI is closed,
inductor current IL passes from the input voltage Vin to the load (V in - Vol appears across inductor.
So IL increases.
When SI is open, stored energy in the inductor forces IL to continue to pass in the load, and
return through diode. The inductor voltage is now reversed and is ::: Vo' So IL decrease.
Switching and Ie Voltage Regulators
8;6
Voltage Multipliers
8.6.1 \
Half Wave Voltage Doubler Circuit
267
During the first negative half oycle, 0 1 is forward biased. So C 1 will get charged, to the peak of the
input voltage Vp, with the polarity as shown in Fig. 8.18. During the positive half cycle, O2 is forward
biased. Therefore, C2 will get charged to Vp + VP = 2V p. Capacitor C 1 will charge C; to Vp and from
the source, through O2, C2 will get charged to a further value of Vp. So after several cycles, the
voltage across C2 will be 2V p. Here RL should be large. Other wise, C2 will be discharging quickly
and the voltage may not be maintained constant at 2V p. This is half wave doubler circuit, since the
output capacitor C2 is charged only once during the full cycle.
Such circuits are used where large output voltage is required. Large Vo can be obtained by
using a bigger transformer. But its cost will be more.
t
O--t'----T---.---
-+t
Fig. 8.18 Half wave voltage double circuit
8.6.2
Full Wave Voltage Regulator
During the positive half ~ycle, 0 1 is forward biased C 1 gets charged to Vp. During the negative
half cycle, O2 is forward bi~ed. So C2 gets charged to Vp. In the steady state, the voltage across RL
in 2 Vp. He~e center tapped transformer is not used. It is called full wave, 'since one of the diodes is
conducting in each half cycle (similar Ito a full wave rectifier circuit). But the problem is, since no
centre tapped transformer is used, there is ho common ground between the input and the output. If the
bottom of RL is grounded, input is floating. This is not desirable in electronic circuits. So in FWR
circuits, centre tapped transformer is used. In this circuit ripple will be less as shown in Fig. 8.19.
Fig. 8.19 Voltage Doubler Circuit
268
8.6.3
Electronic Circuit Analysis
Voltage Tripier
The circuit is as shown in Fig. 8.20. Working is similar to the previous circuit.
...---~_
-j
I
1-:--+- - r - - - - r - - - - : : : : i - h----r+
c,
C,
D,
c,
L -____________~_ _~-
D,
1
~+--~------~---
Fig. 8.20 Voltage tripler circuit
8.6.4
Voltage Quadrupler
The circuit is shown in Fig. 8.21. Is there any limit to the number of sections that can be added to get
large voltage? Theoretically it is possible. But the ripple voltage will get worse as additional sections
are added. (A diode is being put across or in series. So ripple will increase). So these circuits are used
only where very high voltage doubling is to be done. The input voltage itself will be higher.
-
-
I
+
c,
C,
0,
0,
0,
0,
I
Vo
Fig. 8.21 Voltage quadrupler circuit
8.6.5
Peak To Peak Detector
The circuit is as shown in Fig. 8.22.
c,
V
1
t
D\
C
2
He
Vo
j
--+ t
Fig. 8.22 Peak to peak detector
Switching and Ie Voltage Regulators
269
The input sine wave is positively clamped. C 1 - Dl is a positive clamped circuit. D2 - C 2 is
a peak detector circuit. When Dl is forward biased, C 1 gets charged to V p. When D is forward
biased, C2 gets charged to 2V p. So output will measure peak to peak value. RL should be large. If the
input is not symmetrical, positive peak and negative peak are not same, the average value or rms value
measured will not be correct. In such cases, first peak to peak detector is connected and then the DC
meter or AC meter are connected. In certain measurements, peak to peak value is to be determined.
In such cases this type of circuits are used.
Ripple Factors
y Ripple Factor for 'c' filter = 4fj f CR
L
RL
y Ripple Factor for 'L' filter = 4fjroL
.fi
YRipple factor Lc filter
=
3
1
1
x 2roC x 2roL
R
~ 3~
Critical Inductance Lc
3X L
Bleeder Resistor RB = -2-
y fro 1t - filter
[XCI]
X-
.fixc
R
L
Ll
Example : 8.3
Design a power supply using a 1t - filter to give DC output of25V at 100 rnA with a ripple factor not
to exceed 0.01 %. Design of the circuit means, we have to determine L, C, diodes and transformers.
Solution:
Design of the circuit means, we have to determine L, C, diodes and transformer
V
I DC
R = - DC
-=
I
L
Ripple factor
y=
25V
=2500
100mA
X ._
Xc_I
.fi . _c
RL
XLI
V
i_~
VDl'
~rot
Fig. 8.23 Ripple voltage waveform
Electronic Circuit Analysis
270
Xc can be chosen to be = Xci.
y=.fi.
X 2
e
R L .x
This gives a relation between
L1
e and L.
e2 L=y
. There is no unique solution to this .
. Assume a reasonaole value ofL which is commercially available and detennine the corresponding
value of capacitor. Suppose L is chosen as 20 H at 100 rnA with a DC Resistance of 370 n
(of Inductor).
e2 = 1..
or
L
V
oc
e=
rY
VI;
Vy
=V-m
2
I
Vy=~
2fc
Now the transformer voltage ratings are to be chosen.
The voltage drop across the choke = choke resistance x loc.
= 375 x 100 x 10-3 = 37.5V.
Voc=25V.
Therefore, voltage across the first capacitor e in the 1t - filter is
Vc = 25 + 37.5 = 62.5 V.
The peak transformer voltage, to centre tap is
V
Vm = (V c) + -Y. (for e filter)
.
2
'
, IDe
Vy= - 2fc
.
0.1
Vm = 62.5v + 2x50xc
Vrms =
Vrn
~. == 60v
,~2
Therefore, a .transformer with 60 - 0 - 60V is chosen. The ratings of the diode should be,
current of 125 rnA, and voltage = PIV = 2Vm = 2 x 84.6 V = 169.2 V.
Switching and Ie Voltage Regulators
271
Example: 8.4
A full wave rectifier with LC filter is to supply 250 v at 100 m.a. DC. Determine the ratings of the
needed diodes and transformer, the value of the bleeder resistor and the ripple, if ~ of the choke
= 400(2. L = 10 Hand C = 20 IlF.
Solution:
V
DC
R =L
I
DC
RL
250V
=::
o.I
= 2,500n.
For the choke input resistor,
and
1
1tE
[ I+~
R
E .=~
m
2
R
L
_ 1tX250(1+ 400) _
2
2500 - 455V
E
rms
455
= =322 V
.fi
Therefore, the transformer should supply 322V rms on each side of the centre tap. This
includes no allowance for transformer impedance, so that the transformer should be rated at about
340 volts 100 rnA, DC.
The Bleeder Resistance
R
3X L
B
=-
2
L= 10H
2E
2x455
I = __In_=::
= 0.256 A
B
31troL 31t x 377 x 10
Ripple factor
0.47
0.47
Y = 4ro2LC-I = 4x377 2 x20xI0-6
The current ratings of each diode = 0.00413, should be 50 rnA.
272
Electronic Circuit Analysis
Example: 8.S
Design a full wave rectifier to supply current in the range 50 rnA to 100 rnA at 300V with a ripple less
than 5V, using,
1. L section filter
2. 1t - section filter
3. C filter
L
Load
IOOmN300V
(75 ±25)
Fig. 8.24 Circuit for Ex. 8.5
= 50 rnA and
IL(min)
IL(min) =
100 rnA
5V lac
r=-=300 Idc
1.1931
Vac
=V =----r:cdc
.. LC = 35.8, where L is in henrys and C in
LeritIcal2: RL
(:.minimum value of IL
=
~F.
50 rnA, we use RB for 50 rnA only)
2ro s
V
300
=
=6KQ
IL
50xl0-3
R == B
Leritieal =
6K
3 X 21t X 50 = 6.4H
Taking 25% more, because for formula of Leritieal' we use fundamental frequency value,
Leritieal = 8 H
.. We choose
L = 10 H/I00 rnA
C
=
3.58
We choose
4~F/450
Now,
V de
~F
V
_ 2Vm
-1t
-
from the equation,
- Ide
R, where
Switching and Ie Voltage Regulators
~
= Rdiodes
273
+
Rsecondary
+
Rchoko
=
0 + 250 + 200 = 4500
=
2: [300 + 0.1 x 450]
=
3451t
-2- = 541.92 V
V
2
542 = 383 V
=
..fi
rms
.. Transformer used will have 390 - 0 - 390 V/I00 rnA.
Diode current rating
ILI2 = 10012 = 50 + 15% due to variation
57.5 rnA.
Nearest current rating of diode = 100 rnA
.. 100 rnA current rating diodes are used.
=
=
PIV = 2Vm :.PIV rating
=
1084 + 15%
=
1400 V
Using tr - section filter :
RB is not required because there is already a path for discharge current, of capacitor, as 50 rnA is
minimumlL ·
I
,2= 30
also
,2 =
5700
C 1C 2 LR L
I
30
=-
L
Fig. 8.25 With
1t -
section filter
Electronic Circuit Analysis
274
\
,.2
5700
2
C LRL
=
RL(mm) =
~a1culating
300V.
100rnA = 3Kn,
RL(max)::;
300
50
for worst case of supply, or the ,lowest value of RL
Select
5700
1
2
C Lx3K 30
L = 101100 rnA
Using
C
Vm
=
=
6 Kn
=
3 Kn, we have·
:. LC2 = 57.02
:. C2 = 57.02
:. C = 7.55 JlF
= 8J.lF/400 V.
{2~C + RL}
Vde+Ide
= 300+ 100 {
=
=
-6 + 450} x 10-3
1t
2 x 2 x 1t x 50 x 8 x 10
300 + 107.5
407.5 = 408V
Vrn
.J2
Vrms =
=
408
.J2
=
288.5 V
Select transformer as 290 - 0 - 290 V/I00 rnA
Voltage across (:2 = 300 V. If load goes down to 50 rnA the voltage increase, is giv( Vin
= Vd,
+ [2ro:C +RJ
[d,
..
Vde + 354, Ide = 50
:. voltage across C2 in worst case is 354 V.
We select 354 + 15% = 408 V
:. Using C2 of 8 JlF/450,
Voltage across
C 1 = 408 + IL (200 n), where IL = 50 rnA.
= 408 + 10 = 418 V
1t
=
'2
=
541.9 = 542 V
Vm
=
'2
Vde
=
Vm
1t
{300 + 0.1
x
Rs}
1t
'2
{300 + 0.1
{Vde + (0.05 x 450)}
336 V150 rnA + 15% increase
= 386 V
450 V is needed for C
x
45Q}
Switching and Ie Voltage Regulators
275
Example : 8.6
Design a FW circuit using bridge rectifier configuration with two sections ofL-C filter; to give ripple
better than 0.02 %. Inductances available are 100 mH, 100 rnA, (r = 100 ohms).
23:J
50:(
Fig. 8.26 Circuit for Ex. 8.6
For two L - section filter,
ripple factor
(i)
Ji
1
r= 3·(4ro 2LC)2
..
RL
L cntlcal >
- -3
ro
Since
L = 100 mH, f= 50 Hz.
R=3roL
~ 9424.78 n
(ii)
The actual value ofRL will depend on the load itself, and the output voltage required, which
is not specified. We assume that RL = 1 K, and the current required is 30 rnA.
(iii)
Find the value of C.
Since
(iv)
Ji
r
1
= 3 · (4ro 2 LC)2
0.02 Ji
1
2
100 =3·(4xro x.lxC 2 )
For f= 50 Hz, L = .IH
We get, C = 10 uF with a voltage rating depending upon the peak - voltage of the secondary.
Transformer:
The dc resistance of the choke given is 100 ohms.
:. voltage across the capacitor of the first section
30 V + Ide· Rchoke
= 30 V + 30 rnA x 100 ohms
=33 V
:. The voltage at the input of the first section
= 33 V + 40 rnA x 100 ohms
= 33 V + 4 V = 37 V.
=
276
Electronic Circuit Analysis
Note that the current flowing is assumed as 40 rnA instead of 30 rnA to take care of the
capacitor current as well. (Value assumed is slightly on the higher side)
Both capacitors C 1 and C2 can have voltage ratings of 50 V.
Transformer secondary voltage (RMS)
= 37/./2 = 26.2 V
Allowing for two diode voltage drops, we may require the voltage on the secondary
= 26.2 + 2 x .6 = 27.4 V.
This does not take into account the drop due to transformer secondary, which will depend on
the rating of the transformer, but we may assume the same to be about 3 ohms for the transformer
(which will create a drop of 3 ohms x 40 rnA = 120 mY).
We may select transformer with turns ratio 220 : '-7.5, 100 rnA.
(v)
(vi)
The diodes should have rating of minimum 100 rnA, PIV = 60 V.
The bleeder resistance should be around 10 times RL , giving
RB = 10 K
R
6 Vin=vin
¢=
R+RZ
R --eq
R+R Z
Fig. 8.27 For Ex. 8.6
Example : 8.7
Design a zener voltage regulator to supply a load current which varies between 10mA and 25mA at
10V. Input supply voltage available is 20 V ± 10 %.
Solution:
R
200n
Vin
20V± 20%
lOY
Fig. 8.28 Circuit for Ex. 8.7
to
1000n
Switching and Ie Voltage Regulators
277
1. Since the output voltage = 10V, the load resistance, RL , varies from
Vl10mA = 1000 Ohms to
V!25mA = 400 Ohms.
2. Given that
Vin{max) = 20 + 0.01
x
20 = 22 V
Vin(min)=20-0.01 x 20= 18V
Hence,
R=
Vin(min) - Vz
--'----~--
IL(max) + Iz(min)
= (22 - 18) VI (25 - 5) rnA
= 200 Ohms,
assuming Iz(min) to be slightly greater than Iu.
3. The zener is called upon to absorb maximum current when Vin is maximum and at the same
time IL is minimum.
Hence,
Iz (max) + IL(min) = (Vin(max» 1 R.
Iz (max) = {(24 - lO)/200} - lOrnA
= 60 rnA.
The zener diode rating, therefore, should be
Pz (max) = Vz x Iz (max) = 10 V x 60mA
= 600 mW.
We will have to select a zener diode of rating of I W. This higher rating will also take care ofthe
inadvertent opening of the load resistance circuit, in which case, the zener current will be equal to
70rnA and the zener power dissipation will be equal to
P z = lOY x 70mA = 700 mW.
In the above analysis, the dynamic resistar.ce of the zener diode is assumed to be equal to zero
(and, therefore, S = 0, as per equation ). However, even if we. assume Rz = 4 Ohms (say), then
Vz will change with the change in the zener current. The zener current changes from its minimum
value of SmA to a maximum of 60mA. Therefore, Vo = the change in the output voltage,
= (60 - 5)mA x 5 Ohms = 0.275V.
Consequently, the stabilisation factor will be
S = Rl (Rz + R) = 5/205 = 0.024.
Example : 8.8
Design a power supply to give 1A at 12 volts.
Solution:
We shall assume an input voltage supply to be available having twice the output voltage, i.e., say,
25V, with a variation of ± 20%.
The input voltage available, therefore, will be between 20 and 30 volts.
278
Electronic Circuit Analysjs
The circuit, rearranged, will be as shown the Fig. 8.29.
R
I
. 25 V =±20%
v
Fig. 8.29 Circuit for Problem 8.8
1. Let us select, first, the transistor T].
VCE across the transistor
= Vin(max) - Vo
30 - 12 = 18V.
Hence, the power dissipation for the transistor T2 = 18V x IL
18.V x lA = 18W.
We will select EeN 149 as our transistor T2• This transistor has a gain of about 3S.
=
2. The base current of the transistor T] for full variation of the load current, is = lA/3S =
about 30 mAo
The transistor Tj will be supplying the base current of transistor T2• With the voltage across
the transistor Tj being practically the same as that of the transistor T2 , the power dissipation of the
transistor T j will be, neglecting VBE2'
= (30 - 12) V x 30mA
= 18 x 30mA
= S40mW.
We will, therefore have to select a transistor which has this capacity, and also can withstand
30mA. We select transistor SL 100 for the purpose, which has a current gain of about so.
This gives
IBI
=
30mA/SO == 0.6mA.
3. This IBI is the load current for the zener diode, which is operated slightly above Izk of,
say, SmA.
R
Vin(min) - Vz
20 -12
=--Iz(min) + Ib(max) S.6mA
= 8V/S.6mA = 1.42 K - Ohms.
We shall select a standard value of the resistance for R.
Switching and Ie Voltage Regulators
279
The values available near about the calculated values are, 1.2 k and 1.5 k. We shall select 1.2 k,
the lower of the two normally available values, as the higher value will force reduction in Iz (min)'
We will rec~lculate the current Iz (min)' using this value of the resistance.
Iz(min) + IB(max) ,: 8V/1.2k
=
6.67 rnA.
Iz(min) = 6.67mA - 0.6mA = 6.07mA.
Hence,
The power rating of the resistor
=
Iz2 x R
=
(6.07mA)2
x
l.2k
= 45mW.
We shall select a resistance of 114 W capacity.
4. To obtain 12 volts at the output, the zener diode will have to be of
12V + V BE1 + V BE2 = 12 + 0.6 + 0.6
We can adopt the strategy shown in Fig. 8.36.
Ie
=
13.2 V.
I
T,
Fig. 8.30 For Ex. 8.8
5. The power rating of the zener diode :
Vz = 12 V
and
Iz = 6.07 rnA
Hence, pz = 12V x 6.07mA = 80mW. We can select 150m zener. In case there is an open circuit on the load side, the zener current will increase to 6.07 rnA + 0.6 rnA = 6.67 rnA, needing the
zener diode to dissipate power,
= 12V x 6.67 rnA = 80mW.
The zener diode selected can take care of the increased dissipation.
We need the following components to build the circuit.
Transistor T 1 = SL 100 or ECN 100.
TransistorT2 = ECN 149
280
Electronic Circuit Analysis
Example : 8.9
Design 5V, SOOma regulated voltage supply. AC supply is obtained from a 230V : 20V transformer.
Solution:
1.8k 6.2V
23:J
50:1
Full wave
c
1\
(Reifeetive)
Fig. 8.31 Circuit for Ex. 8.9
1. The transformer secondary will have a voltage
= 20 V rms
fi
=
20
=
28.28 V peak.
x
2. We will select a capacitor filter even though the load current is large.
Vav = Vp = 28.28V
Actually this voltage should be equal to
(4jR L C)
Vde= (4jR C+l)
L
(4x50xR
x
Vp
xC)
L - - - - x 28.28
------'=
(4x50xR L xC+l)
Since this supply will have to feed, IL + Ir
giving Ide =
= 1A, neglecting Ir,
O.SA, and hence RL = Vd/1de =
= 5V/0.SA = 10 - Ohms
Substituting this value in the equation (i) we have,
V
=
de
=
6
(2000 x 5000x 10- ) x 28.28V,(2000x5000xl0-6 +1)
10
11
x
28.28::; 25.78V..
3. Selection of the transistors.
Let us select the transistor T 2 .
Vce for the transistor T2 = Yin - Vo
= 25.78 - 5 = 20.78V
..... (i)
Switching and Ie Voltage Regulators
281
Since the load current will have to flow through the transistor T2' the power dissipation capacity
of transistor T2 is
Vee x
IL
= 20.78V x 500mA
=
10.39W.
We select a transistor ECN 149 having Ic(max) = 4A., and maximum power dissipation of
30w. This transistor has minimum hFE of 30.
This will mean that a base current of 500mA/30 = 16.67mA will be required to support the
required load current through the transistor.
Selection of the transistor T j
:
The above base current will be supplied by the transistor T}. This is the collector current of
the transistor T I' Voltage across this transistor is practically the same as that of the transistor T2.
Actually it is
= V in x V0
=
V BE2
25.78 - 5.0 - 0.6
=
19.58 V
-
Since the current flowing through the transistor is = 16.67mA, the power dissipation for this
transistor
= 19.58 x 16.67mA
= 326 mW.
This will necessitate the use of a transistor like SL 100 as transistor T I' This transistor has
hFE(mm) = 50.
4. Selection of zener diode
Transistor T2 base is at a voltage
=
=
Va + Vbe
5 + 0.6 = 5.6V.
Likewise, Tj base should have a voltage of Vb2 + 0.6 = 6.2V.
We can take a zener of 6.2V, or a zener of 5.6V in series with a diode to provide about 6.2V at
the base of Tj •
Resistance
where current
R = (V
de
- V ) R=
bj
tvdc -Vbl )
I
25.78-6.2 19.58
=-f
f
I = fhI +-k
fbi = fLI(hFEj x h FE2 )
= 500mA/(50 x 30) = 0.33 mAo
Hence, R = 19.58V/0.33mA = 1.958 Kn.
'We can select a nominal value of 1.8 Kn.
Electronic Circuit Analysis
282
Recalculating the value of the current passing through it,
We have /= /z (approx.) = (25.78 - 6.2)V/1.8k
=
1O.88mA.
The value of the resistance R is thus 1.8k, and since the current passing through it = 10.88mA,
the power rating of R
= (10.88)2
x
1.8k -Ohms = 213mW.
We can select R as 1.8k -Ohms/ 1I2W rating.
Example : 8.10
Determine the component values for a voltage regulator circuit for 12V at 100mA. The circuit is to
operate from a dc supply of 20 ± 5\'.
Solution:
1. Selection of transistor T1
Voltage across the transistor
= Vm(max) - Vo
= 25 - 12 = 13.
The current through the transistor T I
=
IL(max)
=
100mA.
Therefore, the power dissipation required of the transistor
= 13 V
x
lOrnA = 1.3 Watts
We select a transistor ECN 100 for the purpose, which has
Pd(max) = SWat 25 °C.
Ic(max)
=
0.7A at 25 °C.
hFE(min) =
50.
hfe(typ) = 90.
hie = l.3k
The base current required
IB1(max)
=
IC(max/hfe(mm)
= 100mA/50
=2mA.
2. Selection of the zener diode
Since the output voltage V 0 = 12 Volts, we may select a zener diode of a voltage of 6.8V
(between about 50 and 80% of the output voltage). The power dissipation capacity of the zener
should be around 150m W giving,
12 -6.8
- - = 1.04KQ.
SmA
Switching and Ie Voltage Regulators
283
We select Rz = the nearest available standard value of I k -Ohms, which should have power
dissipation capability of (S.2mA)2 x lk = 27mW.
We, therefore select Rz = lk and of liS watts.
3. Selection of transistor T 2
As a thumb - rule, we select IC2 as about, between 10 and SO% of the IBI depending upon the
load current. Let IC2 = Ima., in our case. The voltage across the tr;ipsistor T2
Vo + VBEI
=
VCE2
=
12 + 0.6 - 6.S
=
S.SY.
=
-
Vz
This means that the Pd(max) of the transistor T2
=
S.SV x ImA = S.S mW.
We select Be 14 7B, which has,
O.2SW
P d(max) =
Ic(max) =
O.IA.
hFE(min) =
hje(mm) =
hie =
200.
240.
4.SKQ
4. Selection of the resistance R3
The maximum base current required for the transistor for thll maximum collector current of-100mA (which in reality is the load current and hence the etl1l.\tter current). We neglect the
additional current required for zener and the base current, making load current = emitter current
= collector current.
IC/hFE(min)
Also the value of
IC2 =
= 100mA/SO = 2.0mA.
1rnA, giving the value of the current
passing through the resistance R3
=
I
+ IBI
= 2mA + ImA = 3mA.
= IC2
Minimum voltage across the resistance R3
= ~n(min) -
Hence,
R3
Vo + VBEI
=
IS - 12 + 0.6
=
3.6 Volts.
=
3.6V I 3mA
=
1.2 KQ
Electronic Circuit Analysis
284 -
Selecting the nearest standard value available, R3 = 470 Ohms, which will make the total
current to 13.6/470 = 2.9mA.
Since lBI is dependant upon the load current and hence, will be 2mA, IC2 will reduce from
1.0mA to 0.9mA.
The power dissipation will be (3mAf
x
4.7K = 39mW.
We select R3 = 4.7K and of lISW.
Likewise, when the input voltage is maximum, i.e., 25V, the current through the resistance R3
will be equal to
(25 - 12 - 0.6)V 1 470 Ohms = 12.4/470
= about 27mA
Since this current of 27mA (less 2mA for the lBI) becomes the collector current for the
transistor T 2, we have Pd(max) for transistor T2 = 5.SV x 25mA = I 45mW. I
This is well within the capacity of the transistor selected, which otherwise would have to be
changed to a higher power transistor.
S. Selection of Ra and Rb
Voltage
V B2
=
V z + V BE2
= 6.S + 0.6
= 7.4 Volts.
Therefore,
r, = R
a
VB2
V
0
7.4
= -12
=062
",
provided that the value of Rb finally selected be sufficiently small as compared to
{hIE2 + Rz (1 + h pE2 )} , so that the shunting effect of the transistor T2 is negligible, i.e., it is assumed
that the current drawn by the Ra and Rb combination is very large as compared to the base current of
transistor T2 , which is
' ;~
= Ib2 = 0.9 1 200 = 4.5 JlA.
,
Ifwe select, Ir as about I rnA, (alternatively, Ir should be chosen, as a thumb - rule, to be about
10% of the load current), we get,
Ra + Rb = Vo 1 Ir
= 121lmA
= 12k - Ohms.
Rb
and
Ra
Hence,
Ra + 0.62 Ra = 12K - Ohms.
=0.62
Ra = 7.4k -qhms
Switching and Ie Voltage Regulators
and
/ 285
Rb = 4.6k
We can select Ra = 6.8k, Rb = 3.9k and connect a potentiometer of lk in-between giving total
resistance
= 6.8 + 3.9k + IK
11.7 K
=
Adjusting the potentiometer, we will be able to adjust the output voltage finely to the
required value.
The power dissipation for
Ra = (lrY x 6.8
~
6.8mW.
=
Similarly, for Rb, Power dissipation = 4.7 mW.
and for the potentiometer
=
ImW
We may select all these components of 1/8 W capacity.
Let us calculate Sv and Ro for the above example.
We have;
where
Rp = 7.4 K in parallel with 4.6 K
=
2.8 K.
R2 may be assumed as about 10 ohms, at the point of operation, and
K = (Rb) / (Ra + Rb)
=
Hence,
S =
v
=
0.383
2.8K + 4.5K + 10(1 + 240)11000
0.47 x 240 x 0.383
----------~----~---
0.225
We have,
Ro =
5.1 ohms.
These are fairly large values.
We can improve upon these, if s9 lesired, by using a darlington pair of transistors in place of
a single transistor T 1. This will help in increasing the value of the resistance R3 since the base current
requirements will be reduced, resulting in smaller value of the voltage - stabilisation ratio. Since the
total gain increases the product of the individual current - gains of the transistors used in the pair, the
value of the Ro will reduce to a considerable extent.
186
8.7
Electronic Circuit Analysis
Uninterrupted Power Supply (UPS)
WlwtJ the, egular AC main line supply fails, the uninterrupted power supply (UPS) unit delivers the
tbe load, The load may be computer or any electronic system etc. The desirable characteristics
of IJPS are :
pOWt:! 10
I, Voltage and frequency of the UPS output must be stable with close tolerances.
2. The output must be of high quality with minimum distortion
3. Must deliver output power without interruption
UPS units are used in :
I. Computer centres/computer systems
2, Instrumentation systems
3, Process-contro I units
4. Communication systems etc.
The block schematic of UPS system is shown in Fig. 8.32.
Static switch unit (I)
Battery
Synch,
transformer
Mains
VV'v
VV'v
SSM
_----=--~ ~-
~
Fig. 8.32 Block schematic of UPS.
UPS consists of a bank of batteries, depending on the output voltage required, static switch unit
for the invertor (SSI), and static switch unit for the mains (SSM), and synchronizing transformer.
8.7.1 Operation
When the mains are functioning, the battery unit is charged. The inverter suppliers power to the load,
through SSI. After the battery unit is fully charged, it floats. When the mains fail, the Invertor
supplies the load current without interruption, by converting the DC voltage it obtains from the
battery into a pulse width modulated AC voltage. It is filtered converted to Sinusoidal wave form and
supplied to load. The inverter output is always synchronized with the mains. If the inverter fails,
power is supplied to load by the mains, by automatically switching on the SSM and switching off the
SST.
287
Switching and IC Voltage Regulators
8.7.2 Inverter Circuit
The circuit,diagram is shown in Fig. 8.33. It is a pulse width modulated Inverter circuit. The circuit
functions on complementary communication method. Whenever SCR T 4 is turned ON T: gets turned
OFF automatically. Similarly when T2 is tuned ON, T3 immediately gets turned OFF and Vicc--versa.
Output is taken between the terminal C and D. The output waveform is shown in Fig. RJ4.
Fig. 8.33 Pulse width modulated inverter circuit.
-t
(
t3
t4
t5
t2
T
0N
T( ON T2 ON T3 ON
3
T2 ON T4 ON T4 ON T2 0N
T(ON
T2 ON
Fig. 8.34 Waveforms.
During the period t} - t:z : SCRs T\ and T2 are conducting. Output voltage follows input DC
voltage.
At t:z: SCR T4 gets triggered, turning OFF T\. Load current passes through T2 and T4 and Diodes
D2 and D4 depending on the direction of load current.
At t3 : SCR T3 gets triggered and turns OFF T 2 . So point 0 becomes +ve with respect to C.
At t 4: SCR T \ gets triggered. The output voltage becomes zero and remains at this level til ts' This
cycle repeats.
Control Circuit: The block schematic of control circuit is shown in Fig. 8.35 The function of the
circuit is to obtain proper firing signals.
Electronic Circuit Analysis
288
Control voltage
J,
50 Hz square
wave generator
1--+
100 Hz square
wave genrator
--+ 100 Hz triangular I-~
wave genrator
PWM
()
()
C)
()
a
b
c
d
ToSCR'E-Pulse amplifier
gate
T
- 0 e&f
Steering circuit
~
Phase shifter
g-J
Fig. 8.35 Block diagram of control circuit.
Wave forms
V
i
--+ t
(a)
V
i
(b)
V
i
(c)
--+ t
V
i
(d)
--+ t
V
i
--+ t
(e)
V
i
--+ t
Fig. 8.36 Output waveform of Dual d Flip Flop (e) & (f).
(t)
Switching and Ie Voltage Regulators
289
The 50 Hzs square wave output is converted into a 100 Hzs signal, which in turn is converted
into a 100 Hzs triangular waveform. This waveform output is compared with the control voltage, by
the comparator to produce a 100 Hzs output pulse. The width of this pulse is proportional to the input
control voltage. This pulse and the basic 50 Hz square wave are fed as clock input and D input to the
Dual D Flop-Flop. The phase shift between the two output wave forms of the Dual D Flop Flop can
be varied by the control voltage. These outputs are amplified and applied through the steering circuit,
as gating signals to the SCR.
8.7.3 Filter Circuit
The output of the inverter circuit is not a pure sinusoidal wave.
The former equation governing the output is,
4V
en == mt Sin
(np)
2 . Cos n(rot)
where n == 1, 3, 5 ...
P == Pulse width in radians
V == Amplitude
ro == Mains radian frequency
The filter circuits are chosen such that the distortion of the sine wave is within the permissable
limits. The desirable characteristics of filter circuits are:
(i)
High output impedance
(iii)
High efficiency
(ii)
(iv)
Low output impedance
Low cast
The different types of filter circuits used are :
1. Low pass filter
2. Low- Pass-with third harmonic filter
3. Series-parallel filter.
c
(a) Low - Pass filter
(b) Low - Pass with third harmonic filter
(e) Series-parallel filter
Fig. 8.37 Different types of filter circuits
Electronic Circuit Analysis
290
8.7.4 Static Switch
The block diagrams used for static switch for inverter (SSI) and static switch for mains (SSM) are
shown in Fig. 8.38.
SSI
SSM
Inverter
output
~
Mains
Fig. 8.38 Static switch-circuit diagram.
When the inverter circuit is functioning it supplies the load current through Ts and T6 . During
positive half cycle, T5 conducts and during negative half cycle T6 conducts. The gating is done by
direct current controlled through an opto coupler. If the inverter circuit stops functioning, it is sensed
and SCRs T7 and T8 are fired immediately. Then the transmission of gate signals to T5 and T6 stops.
When the Inverter becomes active, the load is again, transferred to it by switching ON Ts and T6, and
switching OFF T7 and T 8'
8.8
Switched Mode Power Supplies (SMPS)
For applications in computer systems, Instrumentation panels, electronic control circuits, a DC source
of about 50W and above is usually required. The required features of such a DC power supply are
1. Less ripple
2. Controlled output voltage.
For such requirements, Switching Mode Power 2uppliers,(SMPS) are used. The schematic of
such a SMPS in shown in Fig. 8.39.
AC
input
, Rectifier
circuit
Fig. 8.39 Schematic of Switched Mode Power Supply (SMPS).
Switching and Ie Voltage Regulators
291
The A.C mains input is converted to DC by a rectifier circuit. The output is not controlled at this
stage. The capacitor C l smoothens the output of rectifier circuit, and reduces ripple voltage. The
output voltage of capacitor C l V CI is as shown in Fig. 8.40.
Vel
___________
i
Fig. 8.40 Output wave form of C l'
An electronic switch like a power transistor or MOSFET changes this output to high frequency
A.C Vch as shown in Fig. 8.41. This voltage is chopped between zero and maximum levels, and can
be controlled by the rate of switching.
~t
Fig. 8.41 Output waveform V ch after switching.
This chopped controlled AC voltage is given as input to the LC filter network L2 C 2. This LC filter
smoothens the AC voltage to give DC output volt'lge with very negligible ripple, as shown in Fig. 8.42.
~t
Fig. 8.42 -Output voltage from the LC filter circuit.
This DC voltage is applied to the load. This is the principle of SMPS circuits.
But why to chop the output of capacitor C l and then again convert this chopped voltage to DC
using LC filter? The reasons are :
1. If the capacitor C l it self were to give DC voltage, the value ofC l has to be very very large
to reduce the value of ripple, which is not practicable. The ripple % is inversely proportional
to the val~e of capacitor.
292
Eleetronic Circuit Analysis
2. If the switching circuit is not used to chop the voltage V cl' the output DC voltage V 0 cannot
be controlled. The output voltage magnitude depends on the (ON/OFF) ratio of the switch.
3. The ripple frequency of the voltage after chopping is high. So the values ofL2 and C2 can
be small within the practicable limits. If the ripple frequency is high, filtering is more
effective and low values of Land C can be used as indicated by the equation given below.
The expression for the ratio of output voltage of the filter VOF to its input voltage (V iF)
V
OF - - : : : - -=
V1F
w 2 LC-l
Where 'ro' is the radial frequency of the harmonics. So higher order harmonics are
attenuated more than th~ lower order harmonics.
If ro 2 L2 C 2 »1,
VOF
ViF
= ro 2 LC
or Attenuation of ripple components is inversely proportional to ro 2. i.e., the higher the
switching rate, the lesser the ripple. So the electronic switching circuit is used in SMPS
circuits. The Diode D in the circuit is required to provide a path for the continuous current
n the inductor when the switch is open.
The Inductor L2 in the filter circuit offers zero resistance to DC component because
XL = 0 for DC and large impedance to high ripple AC components (because XL = jroL). The
capacitor C 2 gets charged to the mean value of the chopped voltage.
8.8.1 Single Transistor Switched Mode Power Supply Using Transformer
SMPS circuit is basically a DC to DC converter. For switching action, power transistor or MOSFETs
can be used.
If power transistors are used as switching devices, the switching rate frequency is to be limited to
40KHzs.
IfMOSFETs are used, this frequency can be upto approximately 200 KHzs. The size of the device
can also be small.
By using a transformer in the circuit, the advantages are :
1. Transformer isolation can be achieved between input and output which is essential in
electronic circuits.
2. The range of output voltage levels can be changed, by having two or more secondaries for
the transformer.
293
Switching and Ie Voltage Regulators
Such a circuit diagram is shown in Fig. 8.43.
AC
input
Uncontrolled
rectifier
Control
circuit
Fig. 8.43 SMPS using transformer.
The different types of circuits configurations used for SMPS are :
1. Step-Down or forward or Buck type
2. Step-Up or flyback or Boost type
3. Configuration of (1) and (2) Buck and Boost type.
The circuit for step-Down or forward SMPS is shown in Fig. 8.44.
~s
E~.
V
f
L
D
C
Fig. 8.44 Buck type converter.
The energy is transferred from the source to load. Th value of the load voltage is less than-the
source voltage. When the switch is closed, the core magnetic flux in the inductor rises. The inductor
current rises at a rate such that its voltage L
(:~)
is equal to the voltage difference (Vp -
V~. When
the switch is open, the source current collapses to zero. So the inductor current falls. The mean value
of the output load voltage is given by the ratio of switch ON period to the total time of each cycle.
t)
The duty cycle S = -t- - .
) +t 2
Electronic Circuit Analysis
294
The wave forms are shown in Fig. 8.45.
1
' -_ _..J.I~t
1
1
Fig. 8.45 Waveforms.
8.8.2 Boost converter
This circuit is also known as flyback or step-up converter. The circuit diagram is shown in Fig. 8.46.
In this circuit, the mean value of the load voltage in greater than that of source voltage. When the
switch is closed, current builds up to give stored magnetic every in the inductor. When the switch is
opened, the current transferes to the capacitor and load via the cliode.
Fig. 8.46 Boost converter.
The waveforms are as shown in Fig. 8.47.
f tl1-----'I I1
1
1
1
1
"--
.
~
I
r---.......
I
is
...
1
•
I
, ............
----+Time
~t
l-.!.
----- ~----- -- 10
I
'
=c:::::
ic :
t
IE-- t(
ON
j='"...J::::
)I(
t2~
OFF
Fig. 8.47 Waveforms
Buck and Boost type is the combination of both these circuits.
Switching and Ie Voltage Regulators
295
1. The IC 7908 is a _ _ _ _ _ _ _ Type IC. Output V0
= _ _ _ _ _ __
2.
The IC 7805 is a
Type IC I giving out put of _ _ _ _ _ __
3.
Example of Non-dissipative type Voltage Regulator Circuit is _ _ _ _ _ _ __
4.
What is a regulator ?
5.
What are the types .of regulators ?
6.
What is line regulation?
7.
What factors does the olp voltage of a regulator depend on ?
8.
What is rippe .rejection?
9.
What is the drawback of Zener regulator? How do we avoid it ?
10. Draw the circuit diagram of a series voltage regulator.
11. What is a preregulator ?
12. Why do we go for a switching regulators?
13. Draw the circuit of a voltage multiplier of multiplicatien factor 'n'.
296
Electronic Circuit Analysis
1. Give the Internal Block schematic and Pin Configuration of 723 Voltage Regulator IC.
2. Draw the circuit for 7805 Voltage Regulator IC and explain its working.
Switching and Ie Voltage Regulators
297
1.
Negative Voltage Regulator. Ic'
2.
Positive Voltage Regulator. Ic'
3.
Switching Voltage Regulator.
4.
Regulator gives constant dc voltage irrespective of variations in the input parameters.
5.
Zener regulator
Shunt regulator
Series regulator
Switching regulator.
6.
Ratio of change in output voltage to the change in input voltage.
7.
Vo = f(V i , IL , T)
8.· It is the ratio of output ripple voltage to the input ripple voltage.
9.
It is a fixed voltage regulator. So we go for series voltage regulator.
10. Circuit diagram.
Q,
I,
1L
R,
V,
v::v,
Vo
R,
1
1 1
11. To achieve high stability, there should be a constant current source or a high impedance before
the series pass transistor this is termed as a preregulator.
12. If there are very high frequency fluctuations in the there would be more dissipation in the
series pass transistor. This can be avoided using a switching regulator.
13. Circuit diagram.
c,
c,
D,
D,
..... n stages
"This page is Intentionally Left Blank"
APPENDIX-l
Colour Codes for Electronic Components
RESISTOR COLOUR CODE :
Number of zeros
(except when
Sec;:, ~::::
',?
: ':
u
,
;:g~ To~ernnce
~illnl'C ~
Fourth Band
First Three Bands
Black
Brown
Red
Orange
Yellow
Green
-0
-I
-2
-3
-4
-5
Blue
Violet
Grey
White
Silver
Gold
-6
-7
-8
-9
0.01
0.1
Gold ±5%
Silver ± 10010
None ±20%
CAPACITOR COLOUR CODE:
Colour
digit
}
_<\ecl:m digit
Capacitance in pF
of zeros
nlp,,.,.nt'p
(%)
dcworking
voltage (x 100V)
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Grey
White
Silver
Gold
No Band
Figure Significant Tolerance (0/_)
~
0
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
9
9
om
10
5
0.1
~
APPENDIX
300
INDUCTOR COLOUR CODE:
Color
f",~
P
Decimal point or first digit - + '
Inductance in j.lH Decimal point or second digit ............. /
{
Number of zeros - - - . - / ,
Tolerance (0/0) - - - . /
Significant Tolerance (%)
Figure
0
Black
Brown
1
Red
2
Orange
3
4
Yellow
5
Green
Blue
6
Violet
7
Grey
8
White
9
Silver
Gold
Decimal point
No Band
COLOUR CODE MEMORY AID: W G VIBGYOR BB (WG Vibgyor BB)
Memory aid
Color
Number
Black
Bruins
Relish
Ornery
Young
Greenhorns
I
Blue
Violets
Growing
Wild
Smell
Good
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Grey
White
Silver
Gold
0
1
2
3
4
5
6
7
8
9
0.01
0.1
10010
5%
10
5
:xl
APPENDIX
301
--=>~/ x
W
--9JE)1/4 W
-==f)B )=mcc
\ / 2W
-
)§W""4'
IW
.)))1 } ..
7'
'N )
2W
Fig. A-I.} Relative size of carbon composition resistors with various power ratings
Specifications of Power Transistors
Device
Ic(A) VCEO(V) VCBO(V) hn:MIN Max
'JYpe PD(W)
f T M.H7s
2N6688 NPN
200
20
200
300
~
80
~
2N3442 NPN
117
10
140
160
~
70
0.08
BUX39
120
30
~
120
15
45
8
30
4
40
50
30
-
25
NPN
ECPI49 PNP
Darlington Pair
2N6052
PNP
150
12
100
100
750
-
4
150
12
100
100
750
-
4
,
2N6059
NPN •
APPENDIX-2
Resistor and Capacitor Values
1)'pical Standard Resistor Values (± 10% Tolerance)
n
n
kQ
kQ
kQ
10 .
100
I
10
100
12
120
11
12
120
11
15
150
1.5
15
150
1.5
18
180
1.8
18
180
1.8
22
220
21
22
220
21
2.7
27
270
2.7
27
270
2.7
33
33
330
33
33
330
33
3.9
39
390
3.9
39
390
3.9
4.7
tf7
470
4.7
tf7
470
4.7
5.6
56
560
5.6
56
560
5.6
6.8
68
680
6.8
68
680
6.8
82
820
81
82
820
n
Mn
Mn
10
15
22
APPENDIX
303
Typical Standard Resistor Values (.± 10% Tolerance)
IF
IF
IF
IF
5
50
500
5000
51
510
5100
56
560
J-LF
J-LF
J-LF
J1F
J1F
J-LF
0.05
0.5
5
50
50
5<XX>
5600
0.056
0.56
5.6
56
6000
0.06
62
620
6200
68
680
6800
75
750
7500
J1F
0.068
6
0.68
820
8200
91
910
9100
100
1000
110
1100
120
1200
130
1300
150
1500
160
1600
18
ISO
~
24
6(XX)
6.8
75
8000
82
0.082
0.82
8
II)
82
82
10
100
0.01
0.1
0.012
0.12
12
0.015
0.15
1.5
15
ISO
1800
0.Ql8
0.18
1.8
18
ISO
200
2000
0.02
02
2
~
200
240
2400
250
2500
V
VO
2700
_0.027
027
30
300
3000
0.03
33
330
3300
36
360
3600
39
390
10
12
15
5600
1000 10,000
1500
2000
240
025
25
250
2.7
V
VO
oJ
3
30
300
3000
0.033
0.33
3.3
33
330
3300
3900
0.039
0.39
3.9
39
4000
0.04
43
430
4300
47
470
4700
0.047
4
0.47
4.7
400
47
2500
304
APPENDIX
Physical constants
Charge of an electron
e
1.60 x 10-19 coulombs
Mass of an electron
m
9.09 x 10-31 Kg
elm ratio of an electron
elm
1.759 x 1011 ClKg
Plank's constant
h
6.626 x 10-34 J-sec
Boltzman's constant
K
1.381 x 10-23 JfOK
K
8.62
10-5 ev/oK
6.023 x 1023 molecules/mole
Avogadro's number
Velocity oflight
x
c
3 x,10 8 mlsec
Permeability offree space
1.257 x 10-6 Him
Permittivity offree space
8.85 x 10-12 F/m
Intrinsic concentration in silicon at 300 OK
nj
= 1.5 x 10 19 /cm3
Intrinsic resistivity in silicon at 300 oK
rj
=230,000W--cm
Mobility of electronics in silicon
2
Il\ = 1300cm /V-sec
Mobility of holes in silicon
2
~ =500cm /V-sec
Energy gap at in silicon at 300 oK
= 1.1 ev.
APPENDIX-3
Capacitors
Capacitance
The farad (F) is the Sf unit of capacitance.
The farad is the capacitance of a capacitor that contains a charge of 1 coulomb when the
potential difference between its terminals is 1 volt.
Leakage Current
Despite the fact that the dielectric is an insulator, small leakage currents flow between the plates of a
capacitor. The actual level of leakage current depends on the insulation resistance of the dielectric.
Plastic film capacitors, for example, may have insulation resistances higher than 100,000 MO. At the
other extreme, an electrolytic capacitor may have a microampere (or more) of leakage current, with
only 10 V applied to its terminals.
Polarization
Electrolytic capacitors normally have one terminal identified as the most positive connection. Thus,
they are said to be polarized. This usually limits their application to situations where the polarity of the
applied voltage will not change. This is further discussed for electrolytic capacitors.
Capacitor Equivalent Circuit
An ideal capacitor has a dielectric that has an infinite resistance and plates that have zero resistance.
However, an ideal capacitor does not exist, as all dielectrics have some leakage current and all capacitor
plates have some resistance. The complete equivalent circuit for a capacitor [shown in Fig. A-3.1(a)]
cons:istsofan rlealcapacitorC in series with a resistance RD representing the resistance of the plates,
and in parallel with a resistance RL representing the leakage resistance of the dielectric. Usually, the
plate resistance can be completely neglected, and the equivalent circuit becomes that shown in
Fig. A-3.I(b). With capacitors that have a very high leakage resistance (e.g., mica and plastic film
capacitors), the parallel resistor is frequently omitted in the equivalent circuit, and the capacitor is
then treated as an ideal capacitor. This cannot normally be done for electrolytic capacitors, for example,
which have relatively low leakage resistances. The parallel Rc circuit in,
306
APPENDIX
c
0----1L!J---t_0
I---
RL
(a) Complete equivalent circuit
(b) Parallel equivalent circuit
(c) Series equivalent circuit
Fig. A.3.1
A capacitor equivalent circuit. consists o©f the capacitance C, the leakage resistance RL in
parallel with C, and the plate resistance Ro in series with C and RL "
Fig. A. 3.1 (b) can be shown to have an equivalent series RC circuit, as in Fig. A. 3.I(c). This
is treated in Section 20-6.
A variable air capacitor is made up of a set of movable plates and a set of fixed plates
separated by air.
Because a capacitor's dielectric is largely responsible for determining its most important
characteristics, capacitors are usually identified by the type of dielectric used.
Air Capacitors
A typical capacitor using air as a dielectric is illustrated in Fig. A.3.2. The capacitance is variable, as
is the case with virtually all air capacitors. There are two sets of metal plates, one set fixed and one
movable. The movable plates can be adjusted into or out of the spaces between the fixed plates by
means of the rotatable shaft. Thus, the area of the plates opposite each other is increased I)r decrell and the capacitance value if altered.
Movable
\Plates
Fixed
Fig. A. 3.2 A variable air capacitor is made up of a set of movable plates and a set of fixed
plates separated by air
Paper Capacitors
In its simplest form, a paper capacitor consists of a layer of paper between two layers of metal foil.
The metal foil and paper are rolled up, as illustrated in Fig. A.3.3 (a); external connections are brought
out from the foil layers, and the complete assembly is dipped in wax or plastic. A variation of this is
the metalized paper construction, in which the foil is replaced by thin films of metal deposited on the
surface of the paper. One end of the capacitor sometimes has a band around it [see Fig. A.3.3 (b)].
This does not mean that the device is polarized but simply identifies the terminal that connects to the
outside metal film, so that it can be grounded to avoid pickup of unwanted signals.
APPENDIX
30',
Paper capacitors are available in values ranging from about 500 pF to 50IlF, and in dc working
voltages up to about 600 V. They are among the lower-cost capacitors for a given capacitance value
but are physically larger than several other types having the same capacitance value.
Plastic Film Capacitors
The construction of plastic film capacitors is similar to that of paper capacitors, except that the paper
is replaced by a thin film that is typically polystyrene or Mylar. This type of dielectric gives insulation
resistances greater than 100 000 MO. Working voltages are as high as 600 V, with the capacitor
surviving 1500 V 5urges for a brief period. Capacitance tolerances of ± 2.5% are typical, as are
temperature coefficients of 60 to 150 ppm/oC .
...
Plastic film capacitors are physically smaller but more expensive than papel
are typically available in values ranging from 5 pF to 0.47 1lF.
~apacitors.
They
Foil
Band identifies
outer foil terminal
(a) Construction of a paper capacitor
(b) Appearance of a paper capacitor
Fig. A. 3.3 In a paper capacitor, two sheets of metal foil separated by a sheet of paper are
rolled up together. External connections are made to the foil sheets.
Mica Capacitors
As illustrated in Fig. A. 3.4(a), mica capacitors consist of layers of mica alternated with layers of
metal foil. Connections are made to the metal foil for capacitor leads, and the entire assembly is
dipped in plastic or encapsulated in a molded plastic jacket. Typical capacitance values range from
I pF to O.IIlF, and voltage ratings as high as 35 000 Yare possible. Precise capacitance values and
wide operating temperatures are obtainable with mica capacitors. In a variation of the process, silvered
mica capacitors use films of silver deposited on the mica layers instead of metal foil.
Ceramic Capacitors
The construction of a typical ceramic capacitor is illustrated in Fig. A. 3.4(b). Films of metal are
deposited on each side of a thin ceramic disc, and copper wire terminals are connected to the metal.
The entire units is then encapsulated in a protective coating of plastic. Two different types of
ceramic are used, one of which has extremely high relative permitivity. This gives capacitors that are
much smaller than paper or mica capacitors having the same capacitance value. One disadvantage of
this particular ceramic dielectric is that its leakage resistance is not as high as with other types.
Another type of ceramic gives leakage resistances on the order of 7500 MO. Because of its lower
permitivity, this ceramic produces capacitors that are relatively large for a given value of capacitance.
The range of capacitance values available with ceramic capacitors is typically 1 pF to 0.1 IlF,
with dc working voltages up to 1000 V.
APPENDIX
308
Metal film
~
---
-
Ceramic disk
Mica .- .. ~ ,
(a) Construction of mica capacitor
(b) Ceramic capacitor
Electrolyte
Tantalum
(c) Ceramic trimmer
(d) Construction of tantalum capacitor
Fig. A. 3.4 Mica capacitors consist of sheets of mica interleaved with foil. A ceramic disc
silvered on each side makes a ceramic capacitor; in a ceramic trimmer, the plates area is
screwdriver adjustable. A tantalum capacitor has a relatively large capacitance in a
small volume.
\
Fig. A. 3.4(c) shows a variable ceraffi'ic capacitor known as a trimmer. By means of a
screwdriver, the area of plate on each side of a dielectric can be adjusted to alter the capacitance
value. Typical ranges of adjustment available are 1.5 pF to 3 pF and 7 pF to 45 pF.
Electrolytic Capacitors
The most important feature of electrolytic capacitors is that they can have a very large capacitance in
a physically small container. For example, a capacitance of 5000 IlF can be obtained in a cylindrical
package approximately 5 cm long by 2 cm in diameter. In this case the dc working voltage is only
voltage is only lOY. Similarly, a 1 F capacitor is available in a 22 cm by 7.5 cm cylinder, with a
working voltage of only 3 V. Typical values for electrolytic capacitors range from 1 IlF through
1000001lF.
The construction of an electrolytic capacitor is similar to that of a paper capacitor
(Fig. A.3.5(a)). Two sheets of aluminium foil separated by a fine gauze soaked in electrolyte are
rolled up and encased in an aluminium cylinder for protection. When assembled, a direct voltage is
applied to the capacitor terminals, 'and this causes a thin layer of aluminium oxide to form on the
surface of the positive plate next to the electrolyte (Fig. A.3.5(b)). The aluminium oxide is the
dielectric, and the electrolyte and positive sheet of foil are the capacitor plates. The extremely thin
oxide dielectric gives the very large value of capacitance.
It is very important that electrolytic capacitors be connected with the correct polarity. When
incorrectly connected, gas forms within the electrolyte and the capacitor may explode! Such an
explosion blows the capacitor apart and spreads its contents around. This could have tragic
consequences for the eyes of an experimenter who happens to be closely examining the circuit when
the explosion occurs. The terminal designated as positive must be connected to the most positive of
APPENDIX
309
",/ Foil (+ plate)
Foil
} - (- plate)
'Foil
Electrolyte soaked \ Foil
gauze
(b) The dielectric is a thin layer of
aluminium oxide
(a) Rolled-up foil sheets and electrolytesoaked gauze
Fig. A. 3.5 An electrolyte capacitor is constructed of rolled-up foil sheets separated by electrolytesoaked gauze, the dielectric is a layer of aluminium oxide at the positive plate
the two points in the circuit where the capacitor is to be installed. Fig. A. 3.6 illustrates some circuit
situations where the capacitor must be correctly connected. Nonpolarized electrolytic capacitors can
be obtained. They consist essentially of two capacitors in one package connected back to back, so
one of the oxide films is always correctly biased.
Electrolytic capacitors are available with dc working voltages greater than 400 Y, but in this
case capacitance values do not exceed 100 mF. In addition to their low working voltage and polarized
operation. Another disadvantage of electrolytic capacitors is their relatively high leakage current.
,------D+ 12 V
.------~+12V
, I'
RI
+5Vt-----,
+5V
(a) Capacitor connected between +5 V and ground
+---------..
(b) Connected between +7 V and +5 V
,-------~
+ 12 V
1----~5V
ac
voltage
(c) Connected between + 5.7 V and a grounded ac voltage source
Fig. A. 3.6 It is very important that polarized capacitors be correctly connected . The capaCitor
positive terminal voltage must be more positive than the voltage at the negative terminal.
310
APPENDIX
Tantalum Capacitors
This is another type of electrolytic capacitor. Powdered tantalum is sintered (or baked), typically into
a cylindrical shape. The resulting solid is quite porous, so that when immersed in a container of
electrolyte, the electrolyte is absorbed into the tantalum. The tantalum then has a large surface area
in contact with the electrolyte (Fig. A. 3.5). When a dc forming voltage is applied, a thin oxide film
is formed throughout the electrolyte-tantalum contact area. The result, again, is a large capacitance
value in a small volume.
Capacitor Color Codes
Physically large capacitors usually have their capacitance value, tolerance and dc working voltage
printed on the side of the case. Small capacitors (like small resistors) use a code of colored bands
(or sometimes colored dots) to indicate the component parameters.
There are several capacitor color codes in current use. Here is one of the most common.
l
First digit
~~iii9~ -==:::::::= Second digit
Capacitance in pF
-+- - Number
of zeros
'-..-,- ---,...... __._..__ Tolerance (%)
dc working voltage ( x 100 V)
Color
Black
Brown
Red
Orange
YeIIow
Green
Blue
Violet
Grey
White
Silver
Gold
No band
ure
Si
0
I
2
3
4
5
Tolerance (%)
20
2
3
4
5
6
7
6
8
9
0.01
0.1
8
7
9
10
5
20
A typical tantalum capacitor in a cylindrical shape 2 cm by I cm might have a capacitance of
100 mF and a dc working voltage of20 V. Other types are available with a working voltage up to 630
V, but with capacitance values on the order of 3.5 mF. Like aluminium-foil electrolytic capacitors,
tantalum capacitors must be connected with the correct polarity size of the inductor, the maximum
current can be anything from about 50 mA to I A. The core in such an inductor may be made
adjustable so that it can be screwed into or partially out of the coil. Thus, the coil inductance is
variable. Note the graphic symbol for an inductor with an adjustable core [Fig. A. 3.6(b)].
APPENDIX-4
Inductors
Magnetic Flux and Flux Density
The weber* (Wb) is the Sf unit ofmagneiicflux.
The weber is defined as the magnetic flux which, linking a single-turn coil, produces an emfof
1 V when the flux is reduced to zero at a constant rate in 1 s.
The tesla*" (T) is the Sf unit of magnetic flux density.
The tesla is the flux density in a magnetic field when 1 Wb offlux occurs in a plane of 1 m2;
that is, the tesla can be described as 1 Wblm2.
Inductance
The Sf unit of inductance is the henry (H).
The inductance of a circuit is 1 henry (l H) when an emf of 1 V is induced by the current
changing at the rate of 1 Als.
Molded Inductors
A small molded inductor is shown in Fig. A. 4.2(c). Typical available values for this type range from
1.2
~H
to 10 mH, maximum currents of about 70 rnA. The values of molded inductors are identified
by a color code, similar to molded resistors. Fig. A. 4.2(d) shows a tiny-film inductor used in certain
types of electronic circuits. In this case the inductor is simply a thin metal film deposited in the form
of a spiral on ceramic base.
Laboratory Inductors
Laboratory-type variable inductors can be constructed in decade box format, in which precision
inductors are switched into or out of a circuit by means of rotary switches. Alternatively, two coupled
coils can be employed as a variable inductor. The coils may be connected in series or in parallel, and
the total inductance is controlled by adjusting the position of one coil relative to the other.
312
APPENDIX
Color Code For Small Inductors
~~.,
<o-~R""____
--:.-..-~~~..,
~ '~
Decimal point or fir:
Indn'tan" ;n.H
f D,dm'! p,;nt ' " ' '
1
Number of zeros
Tolerance (%)
Color
Significant Figure
Black
0
Brown
I
Red
2
Orange
3
Yellow
4
Green
5
Blue
6
Violet
7
Grey
8
White
9
Silver
Gold
Tolerance (%)
10
decimal point
No band
"
ffi
. \
~)
'
,
5
20
0
"
Coil
~
' ,-"
Bobb In
Ferrite pot core
/
tI
Fig. A. 4.1 Some low-current, high-frequency inductors are wound on bobbins contained in a ferrite pot core. The ferrite core increases the winding inductance and screens the inductor
Coil to protect adjacent components against flux leakage and to protect the coil from external
magnetic fields. The coil is wound on a bobbin, so its number of turns is easily modified.
Three different types of low-current inductors are illustrated in Fig. A. 4.2. Fig. A 4.2(a)
shows a type that is available either as an air-cored inductor or with a ferromagnetic core. With an air
core, the inductance values up to about 10 mH can be obtained. Depending on the thickness of wire
used and the physical.
APPENDIX
313
J\
II
II
1/
II
II
II
(a) Inductor with air core or.
ferromagnetic core
(c) Molded inductor
(b) Circuit symbol for an inductor with
an adjustable ferromagnetic core
(d) Thin-film inductor
Fig. A. 4.2 Small inductors may be wound on an insulating tube with an adjustable ferrite core,
molded like small resistors, or deposited as a conducting film on an insulating material
If the mutual inductance between two adjacent coils is not known, it can be determined by
measuring the total inductance of the coils in series-aiding and series-opposing connections. Then,
L=L+L+2M
for series-aiding
and
L=L+L-2M
for series-opposing
Subtracting,
L -L
a
1
b
2
1
a
b
2
=
4M
Therefore,
M
=
k~LJL2
From these two equations, the coefficient of coupling of the two coils can be determined.
Stray Inductance
Inductance is (change in flux linkages) / (change in current). So every current-carrying conductor
has some self-inductance, and every pair of conductors has inductance. These stray inductance are
usually unwanted, although they are sometimes used as components in a circuit design. In dc
applications, stray inductance is normally ~nimportant, but in radio frequency ac circuits it can be
considerable nuisance. Stray inductance is normally minimized by keeping connecting wires as short
as possible.
APPE~IX
314
- - - - - - - - - - - - - Summary of Formulae - - - - - - - - - - - - ~<I>
Induced emf
e
Induced emf
e =--
Inductance
L=~
Inductance
L=--
Flux change
~<I>
=
Self - inductance
L
II II
r·
ro ~i JtI2 -t
Mutual inductance
M=~
=-
M
L
~ .cI>
M
L
~i/M
~<I>N
~i
=
A
N -t
.
A
~i
II II
r'
ro
~i/ ~t
~<I>N_,_
Induced emf
e
Mutual inductance
M= ~<I>N,
Mutual inductance
M= k ~<I>Ns
Mutual inductance
M= k~LIL2
Energy stored
W=
Energy stored
B2AI
W=--
=-~
L
ill
ill
1
-
2
L/2
2110
_ IrJductances in series
L
s
=
L + L +-L + .....
I
2
3
1
1
1
1
Lp
Ll
L2
~
Inductances in parallel
- = - + - + - + .....
Total inductanc/! (series-aiding)
L=L +L +2M
Total inductance (series-opposing)
L=L +L -2M
Mutual inductance
Lo -Lb M=----"'4
I
I
2
2
APPENOIX-5
Miscellaneous
Ionic Bonding
In some insulating materials, notably rubber and plastics, the bending process is also covalent. The
valence electrons in these bonds are very strongly attached to their atoms, so the possibility of
current flow is virtua1ly zero. In other types of insulating materials, some atoms have parted with
outer-shell electrons, but these have been accepted into the orbit of other atoms. Thus, the atoms are
ionized; those which gave up electrons have become positive ions, and those which accepted the
electrons become negative ions. This creates an electrostatic bonding force between the atoms,
termed ionic bonding. Ionic bonding is found in such materials as glass and porcelain. Because
there are virtually no free electrons, no cur~ent can flow, and the material is an insulator.
histilators
Fjg. A. 5.1 shows some typical arrangements of conductors and insulators. Electric cable usually
consists of conducting copper wire surrounded by an insulating sheath of rubber or plastic. Sometimes
there is more than one conductor, and these are, of course, individually insulated.
Conductor - Insulation --.....
Printed copper
conductor
Insulation
Protective _ _
sheath
Fig. A. 5.1 Conductors employed for industrial and domestic purposes normally have stranded
copper wires with rubber or plastic ins.ulation. In electronics equipment, flat cables of fine wires
and thin printed circuit conductors are widely used.
Conductors
The function of a conductor is to conduct current form one point to another in an electric circuit. As
discussed. electric cables usually consist of copper conductors sheathed with rubber or plastic
316
APPENDIX
Screened
coaxial
cable
Conductor
Screen
Circular
multiconductor
cable
Flat
multi conductor
cable
Fig. A. 5.2 Many different tYpes of cables are used with electronics equipment:
insulating material. Cables that have to carry large currents must have relatively thick conductors.
Where very small currents are involved, the conductor may be a thin strip of copper oreven an
aluminium film. Between these two extremes, a wide range of conductors exist for various applications.
Three different types of cables used in electronics equipment are illustrated in Fig. A. 5.2 conductor
and a circular plaited conducting screen, as well as an outer insulating sheath. The other two are
multiconductor cables, one circular, and one flat.
.
Because each conductor has a finite resistance, a current passing through it causes a voltage
drop from one end of the conductor to the other (Fig. A. 5.3). When conductors are long andIor
carry large currents, the conductor voltage drop may cause unsatisfactory performance of the equipment
supplied. Power (12 R) is also dissipated in every current-carrying conductor, and this is, ofcourse,
wasted .power;
f~........
~1
E. . . . . .
~
I
(a) Current flow through a conductor produces a voltage drop along the conductor
f:E~
o----~~~-~~-------o
I
R'-.
Conductor resistance
(b) Conductor resistance causes voltage drop when a current flows
Fig. A. 5.3 Conductor resistance (R) is determined by applying the voltage drop and current
fevel to Ohm's law. The resistance per unit length (RlI) is then used to select a suitable
wire gauge.
APPENDIX
317
Porcelain coating
Color bands to;;.-;;;;::;:=:::e
identify resistance
value
Carbon ,",UJJJI',};"lJUJJ
resistance element
wire
Connecting wires
(a) Wire-wound resistor
(b) Carbon composition resistor
Ceramic base
/
Metal film
t
Protective insulating cover
(c) Metal film resistor
Fig. A.5.4 Individual resistors are typically wire-wound or carbon composition construction.
Wirewound resistors are used where high power dissipation is required. Carbon composition
type is the least expensive. Metal film resistance values can be more accurate than carbon
composition type.
The illustration in Fig. A. 5.5(a) shows a coil of closely wound insulated resistance wire
formed into partial circle. The coil has a low-resistance terminal at each end, and a third terminal is
connected to a movable contact with a shaft adjustment facility. The movable contact can be set to
any point on a connecting track that extends over one (unisulated) edge of the coil. Using the adjustable
contact, the resistance from either end terminal to the center terminal may be adjusted from zero to
the maximum coil resistance.
Another type of variable resistor, known as a decade resistance box, is shown in
Fig. A. 5.5(c). This is a laboratory component that contains precise values of switched seriesconnected resistors. As illustrated, the first switch (from the right) controls re~istance values in
10 steps from on to 90 and the second switches values of 100,200,300, and so on. The decade
box shown can be set to within + 10 of any value from 00 to 99990. Other decade boxes are
available with different resistance ranges.
Resistor Tolerance
Standard (fixed-value) resistors normally range from 2.70 to 22MO. The resistance tolerances on
these standard values are typically ± 20%, ± 10%, ± 5% or ± 1%. A tolerance of + 10% on a 1000
resistor means that the actual resistance may be as high as 1000 + 10% (i.e., 1100) or as low as
WOO - 10% (Le., 900). Obviously, the resistors with the smallest tolerance are the most accurate
and the most expensive.
318
APPENDIX
Moving
End
Shaft for
adjustment
contact
terminal
(8) Typical construction of a resistor variabie resistor (and potentiometer)
1
(b) Circuit symbols for a variable resistor
(c) Decade resistance box
Fig. A. 5.5 Small variable resistors are used in electronic circuit construction. Large de~de
, resistance boxes are employed!n electronics laboratories. "
APPENDIX
319
More Resistors
14 pin dual-in-line
package
•
Internal
resistor
arrangement
Resistor Networks
Resistor networks are available in integrated circuits type dual-in
lin package. One construction method uses a thick film techniqu
in which conducting solutions are deposited in the required form.
Photoconductive Cell
This is simply a resistor constructed of photoconductive material
(cadmium selenide or cadmium sulfide). When dark, the cell
resistance is very high. When illuminated, the resistance decreases
in proportion to the level of illumination.
Resistance
Contact
wire
ajusting screw
....
~~
wrre '/ r \ \
/
Slidi'lg
contact
;' ___ ~ __ -...-_~ .. y.".~l
Low Power Variable Resistor
A small variable resistor suitable for mounting directly on a cireui
board. A threaded shaft1 which is adjustable by a screwdriver, se
the position of the moving contact on a resistance wire.
Wire terminals
High Power Resistor
High power resistors are usually wire-wound on the surface of a
ceramic tube. Air flow through the tube helps to keep the resistor
from overheating.
320
APPENDIX
Two memory aids for determining the direction of the magnetic flux around a current-carrying
conductor are shown in Fig. A. 5.6. The right-hand-screw rule as illustrated in Fig. A. 5.6(a) shows
a wood screw being turned clockwise and progressing into a piece of wood. The horizontal direction
of the screw is analogous to' the direction of current in a conductor, and the circular motion of the
screw-shows the direction of magnetic flux around the cond,uctor. In the right-hand rule, illustrated
in Fig. A. S.6(b), a right hand is closed around 'a conductor with the thumb ointing in the
(conventional) di~ction of <;:urrent flow. The fingers point in the direction of the magnetic lines of
fotce around the conductor.
Because a current-carrying conductor has a magnetic field around it, when two current-carrying
conductors are brought close together there will be interaction between the fields. Fig. A. 5.7(a)
shows the effect, on the fields when two conductors carrying in opposite direction are adjacent. The
directions of the magnetic passes through the center of the coil. Therefore, the one-turn coil acts like
a little magnet and has a magnetic field \with an identifiable N pole and S pole. Instead ofa single turn,
the coil may have many turns, as illustrated in Fig. A. 5.7(c). In this case the flux generated by each
of the individual current-carrying turns tends to link up and pass out of one end of the coil and back
into the other end. This type of coil, known as a solenoid, obviously has a magnetic field pattern very
similar to that of a bar magnet.
Direction of
Direction
rotation
of motion
Direction of lines offorce (same as
.
f
'
f
)
'
d lrectlOn
0 rotation 0 screw
C
d t
on uc or
~
3@0;8 (9,
Cuneo! direc:iOO
(same as direction of
motion of screw)
(a) Right-hand screw rule
Thumb points in
current direction
I
-+-
Conductor
Fingers point in direction
of lines offorce
(b) Right-hand rule
Fig. A. 5.6 The right-hand-screw rule and the right-hand rule can be used for determining the
direction of the magnetic lines of force around a current-carrying conductor.
321
APPENDIX
The right-hand rule for determining the qirection of flux from a solenoid is illustrated in
Fig. A. S.7(d). When the solenoid is gripped with the right hand so that the fingers are pointing in the
direction of current flow in the coils, the thuplb points in the direction of the flux (Le., toward the Npole end of the solenoid).
Electromagnetic Induction
It has been demonstrated that a magnetic flux is generated by an electric current flowing in a conductor.
The converse is also possible; that is, a magnetic flux can produce a current flow in a conductor.
"-
"
---- ----Flux around
conductor
"
~
(a) All of the flux passes through the center of
the coil turn
--- ------------
I~
.... - --,
" .
'
.....
~'"
(b) Side view of coil turn and flux
-~-
,i
(c) A soler1oid sets up a flux like a bar
magna
Fingers point in
current direction
(d) Right-hand rule for solenoid flux
direction
Fig. A. 5.7 In current-carrying coils, the magnetic lines of force around the conductors all pass
through the center of the coil.
322
APPENDIX
Consider Fig. A. S.8(a), in which a handled bar magnet is shown being brought close to a coil of wire.
A:; ~e bar magnet approaches the coil, the flux from the magnet brushes across the coil conductors
or cuts the condJ.1ctors. This produces a current flow in the conductors proportIonal to the total flux
that cuts the coil. If the coil circuit is closed by a resistor (as shown broken in the figure), a current
flows. Whether or not the circuit is closed, an electromotive force (eDit) can be measured at the coil
,terminals. This effect is known as electromagnetic induction.
Flux brushe.
/ over coil turns
(a) emf induced in a coil by the motion of the flux from the bar magnet
(b) emf induced in a coil by the motion of the flux from the solenoid when the current is switched on or off
Fig. A.5.8 An electromotive force (emf) is induced in a coil when the coil is brushed by a
magnetic field. The magnetic field may be from a bar magnet or from a current-carrying coil.
323
APPENDIX
-Flux
N
(a) Single-turn coil pivoted in a magnetic field
' .... _-+
--+-- ....
s
\
..... -~-.".
(b) Showing the force on each side of a single-turn pivoted in a magnetic field
Fig. A. 5.9 A force is exerted on each side of a current-carrying coil pivoted in a magnetic field.
This force tend~ cause the coil to rotate
324
APPENDIX
Fibre Optic Cables :
Light and Other Rays
Light ·rays or waves are a type of energy called electromagnetic
energy. \ They shine out, or radiate, from their source, so they
are called electromagnetic radiatio~ J,ight rays are just a tiny
part of a huge range of rays a'Rd waves called the
electromagnetic spectrum (EMS)
•
The parts of the electromagnetic
spectrum have different wavelengths
and different names.
Radio waves
Micro-waves
Infra-red waves
Light waves
Ultra-violet
waves
X-rays
Gamma rays
Twisted steel centre
cable
Optical fibres are colour-coded, so
that they can be correctly connected
to other machines .
•
Light shines from
optical fibres, making
their tips glow.
protects from dirt,
damp or damage.
Glass cladding
Sheath
Laser light
APPENDIX-6
Circuit Symbols
1
1
-
1
I
dc voltage source
or single-cell
battery
Multicell
battery
Generator
Current
source
t@l
Voltmeter
Ammeter
ac voltage
source
Lamp
11
Chassis
Ground
+++
Fuse
Conductor connection
or junction
Wattmeter
Unconnected crossing
conductors
APPENDIX
326
dc voltage source
or single-cell
battery
1
T
Capacitor
Two-way
switch
Double-pole
switch
f ,
Variable
capacitor
Resistor
Air-cored
inductor
Variable
resistor
,II
Iron-cored
inductor
Potentiometer
+
Variable
inductor
APPENDIX-7
Unit Conversion Factors
The following factors may be used for conversion between non-SI units and SI units.
To Convert
To
Multiply By
Area Units
acres
acres
circular mils
square feet
square inches
square miles
square miles
square yards
square meters (m2)
hectares (ha)
square meters (m2)
square meters (m2)
square centimeters (cm 2)
hectares (ha)
square kilometers (km 2)
square meters (m2 )
4047
0.4047
5.067 x 10-10
0.0909
6.452
259
2.59
0.8361
Electric- and Magnetic Units
amperes/inch
gauses
gilberts
lines/sq. inch
Maxwells
mhos
Oersteds
Energy and Work Units
Btu
Btu
ergs
ergs
foot-pounds
foot-pounds
amperes/meter (Aim)
teslas (T)
ampere (turns) (A)
teslas (T)
webers (Wb)
Siemens(S)
amperes/meter
39.37
10-4
0.7958
1.55 x 10-5
10--8
joules(J)
kilowatt-hours (kWh)
joules (J)
kilowatt-hours (kWh)
joules (J)
kilogram meters (kgm)
1054.8
2.928 x 10-4
10-7
0.2778 x 10-13
1.356
0.1383
1
79.577<
APPENDIX
328
Force Units
dynes
dynes
pounds
poundals
grams
1.02 x 10-3
10-5
grams (g)
newtons (N)
newtons (N)
newtons (N)
newtons(N)
4.448
0.1383
9.807 x 10-3
lumens/cm2
10.764
Illumination Units
foot-candles
To Convert
Multiply By
To
Linear Units
1 x 10-10
meters (m)
meters (m)
meters (m)
centimeters (cm)
meters (m)
kilometers (km)
kilometers (km)
centimeters (cm)
meters (m)
0.3048
1.8288
2.54
10""6
1.853
1.609
2.54 x 10-3
watts (W)
745.7
atmospheres
kilograms/sq. meter (kg/m2)
10332
atmospheres
kilopascals (kPa)
kilopascals (kPa)
101.325
angstroms
feet
fathoms
inches
microns
miles (nautical)
miles (statute)
mils
yards
0.9144
Power Units
horsepower
Pressure Units
bars
b¥s
pounds/sq. foot
pounds/sq. ineh
kilograms/sq .meter (kg/m
2
2
)
100
1.02 x 10-4
kilograms/sq .meter (kg/m )
kilograms/sq. meter (kg/m 2)
4.882
degrees Fahrenheit (OF)
degrees celsius (0C)
(OF - 32)/1.8
degrees Fahrenheit (OF)
degrees kelvin (K)
273.15 + COF-32)11.8
703
Temperature Units
329
APPENDIX
Gauge
36
37
38
39
40
Velocity Units
mileslhour (mph)
knots
kilometerslhour (km/h)
kilometerslhour (kmlh)
1.609
1.853
Volume Units
bushels
cubic feet
cubic inches
cubic inches
cubic yards
gallons (U.S.)
gallons (imperial)
gallons (U.S.)
gallons (imperial)
gills
pints (U.S.)
pints (imperial)
cubic meters (m3)
cubic meters (m 3)
cubic centimeters (cm3 )
liters (1)
cubic meters (m 3)
cubic meters (m 3)
cubic meters (m3 )
liters (1)
liters (1)
liters (1)
liters (1)
liters (1)
0.03524
0.02832
16.387
0.01639
0.7646
3.7853 x 10-3
4.546 x 10-3
3.7853
4.546
0.1183
0.4732
0.5683
Copper Wire
Resistance
(Q/km)
Diameter (mm)
0.127
0.113
0.101
0.090
0.080
To Convert
quarts (U.S.)
quarts (imperial)
1360
1715
2147
2704
3422
To
liters (1)
liters (1)
Diameter (mil)
Copper Wire
Resistance
(Q/km)
415
523
655
832
1044
5
4.5
4
3.5
3.1
Multiply By
0.9463
1.137
Weight Units
ounces
pounds
tons (long)
tons (short)
grams (g)
kilograms (kg)
kilograms (kg)
kilograms (kg)
The siemens* is the unit of conductance.
conductance
1
resIstance
= --.- -
2~.35
0.45359
1016
907.18
APPENDIX_\: 8
American Wire Gauge Sizes and
Metric Equivalents
Gauge
Diameter (mm)
Copper wire
Resistance (!lIkm)
Copper wire
Diameter (mill) Resistance (!lit 000 ft)
r
0000
11.68
0.160
460
0.049
000
10.40
0.203
409.6
0.062
00
9.266
0.255
364.8
0.078
0
8.252
0.316
324.
0.098
1
7.348
0.406
289.3
0.124
2
6.543
0.511
257.6
0.156
3
5.827
0.645
229.4
0.197
4
'5
5.189
0.813
204.3
0.248
4.620
1.026
181.9
0.313
6
4.115
1.29
162
0.395
7
3.665
1.63
144.3
0.498
8
3.264
2.06
128.5
0.628
9
2.906
2.59
114.4
0.192
10
2.588
3.27
101.9
0.999
11
2.30
4.10
90.7
1.26
12
2.05
5.20
80.8
1.59
13
1.83
6.55
72
2
14
1.63
8.26
64.1
2.52
.15.
1.45
10.4
57.1
3.18
16
1.29
13.1
50.8
4.02
17
1.15
16.6
45.3
5.06
18
1.0~
21.0
40.3
6.39
331
APPENDIX
19
0.912
26.3
35.9
8.05
20
0.813
33.2
32
10.1
21
0.723
41.9
28.5
12.8
22
0.644
52.8
25.3
16.1
23
0.573
66.7
22.6
20.3
24
0.511
83.9
20.1
25.7
25
0.455
106
17.9
32.4
26
0.405
134
15.9
41
27
0.361
168
14.2
51.4
28
0.321
213
12.6
64.9
29
0.286
267
11.3
81.4
30
0.255
337
10
103
31
0.227
425
8.9
130
32
0.202
537
8
164
33
0.180
676
,7.1
206
34
0.160
855
6.3
2611
35
0.143
1071
5.6
329
Index
333
Index
A
Double tuned 190, 204, 205, 208, 209, 211
Double Tuned Amplifier 191
Amplifiers 2, 21, 31
common base 30,32,45,73,74,217
common collector 30, 32, 45
common emitter 39,49,51,54,67, 73
casco de 33, 72, 85
bandwidth 45, 49, 51
darlington pair 65, 66, 67, 87
emitter follower l3,27, 141, 175
Amplitude 3,35,85,87,88
E
Emitter Bypass Capacitor 53
Emitter Capacitance (Ce ) 99
F
Feedback Conductance 94, 118
FET 24
FET Tuned R.F Amplifier 218
B
G
Gain 214,216
Bandwidth of Amplifiers 35
BJT 3, 146, 147, 209,210, 213
21~
215, 216, 234, 236
Bypass capacitor 53, 54, 78, 85, 155
- current 217
- power 214
- voltage 216
Gain - Bandwidth Product 51
c
H
Cascaded amplifier 43, 48, 49
Heat sinks 176, 180
Class A Operation 146
Hybrid -1t model 98, 120, 123
Class B Operation 147
Class C Operation 147
Collector Capacitance (Cc ) 100
I
Impedance matching
154, 156, 158, 184, 187, 197, 216, 226
Common Base Amplifiers 3 1
J
Complimentary symmetry 168, 169, 183, 186
Conversion efficiency 30, 146, 151, 160, 184
JFET 3, 2l3, 236
Coupling capacitor 4, 34, 223
L
Current Gain 7, 11, 15, 43,44,45, 49, 100
Line Regulation 249, 253, 254
Cut-off frequency 33
Line Regulation 253
D
M
Decibel 33, 54, 60, 61
Mid frequency range 39,62, 84
Direct coupled amplifier 153
Midband frequency 49,58
Miller's Theorem 55
Distortion 35,84,88, 146, 148, 167
Electronic Circuit Analysis
334
N
Shunt Compensation 222, 224
n - Stage Cascaded Amplifier 43
Si,ngle tuned 185, 190
Negative Voltage Regulators 240
Single Tuned Amplifier 191
o
Small Signal Amplifiers 147
Specifications 214, 240, 242, 266
Output Conductance (h22 ) 74
Stabilization 158, 244, 254
Output resistance 54, 69
Stiff Coupling 58, 62, 63
p
Switching regulators 241, 246, 253
T
Phase Inverters 170
Phase Response 51
Tank circuits 218, 220
Power amplifiers 35, 148, 158, 183
Temperature coefficient 240, 244, 254
Power Gain 8, 12,42,43,60
Transconductance 25, 26, 90, 92, 99, 117
Pre regulator 253,254
Transformer coupled amplifier 151, 153, 157
Push-pull amplifier 148, 168
Tuned Circuit Amplifier 34
Q
Q-point 177, 178
/
v
Voltage gain 5, 7, 11,23
R
R-C Coupled Amplifier Circuit (Single Stage) 38
Ripple Rejection 240, 249, 253
s
Series Peaked Circuit 224, 225
Series Voltage Regulator 245,247,253
Voltage quadrupler 270
w
Wideband Amplifiers 221
z
Zener Voltage Regulator 241, 278